4751
Mark Scheme
June 2005
Mark Scheme 4751
June 2005
4751
Mark Scheme
June 2005
Section A
1
40
2
6y
as final answer
3+ m
n + 1 and n + 2 both seen
3n + 3
3
=3(n + 1) o.e.
A1
4
−0.6 o.e.
2
5
(4, 0)
(0, 12/5) o.e.
8 − 12x + 6x2 − x3 isw
1
1
4
6
(i) 1
1
(ii) a8 cao
1
2
3
[ x =]
(iii)
7
8
9
1
or 13 a −3b −1 isw
3a 3b
1
M1
3
(i) 3√6 or √54 isw
2
(ii) 10 + 2√7
3
x (30 − 2x) = 112
x(15 − x) = 56 or 30x − 2x2 = 112
M1
A1
(x − 7)(x − 8)
x = 7 or 8
7 by 16 or 8 by 14
1
1
1
[y=] 3x + 2 = 3x2 − 7x + 1
M1
[0 =] 3x2 − 10x − 1 or −3x2 + 10x + 1
10 ± 100 + 12
x=
6
M1
M1
=
10 ± 112
5 ± 28
or
o.e. isw
6
3
A2
M1 subst of 3 for x or attempt at long
divn with x3 − 3x2 seen in working; 0 for
attempt at factors by inspection
2
M1 for 3x + mx = y + 5y o.e. and
M1 for x(3 + m) or ft sign error
3
condone e.g. a instead of n for last 2
marks or starting again with full method
for middle number = y etc
or 3 a factor of both terms so divisible by 3
3
M1 for 0.6 or −0.6x o.e. or rearrangement
to ‘y =’ form [need not be correct]
condone values of x and y given
4
B3 for 3 terms correct or all correct
except for signs; B2 for two terms correct
including at least one of − 12x and 6x2;
4
B1 for 1 3 3 1 soi or for 8 and − x3
M2 for two ‘terms’ correct or M1 for
1
1
or
3a3b or
; ignore ±
1
6 2
6 2 2
9
a
b
(9a b )
M1 for √(4×6) or 2√6 or 3√2√3seen
5
M1 for attempt to multiply num. and
denom. by 5 + √7 and M1 for 18 or 25 −
7 seen
allow M1 for length = 30 − 2x soi
NB answer given
5
0 for formula or completing sq etc
must be explicit; both values required
allow for 16 and 14 found following 7
and 8; both required
or rearrangement of linear and subst for x
in quadratic attempted
condone one error; dep on first M1
attempt at formula [dep. on first M1 and
quadratic = 0]; M2 for whole method for
completing square or M1 to stage before
taking roots
A1 for two of three ‘terms’ correct [with
correct fraction line] or for one root
5
5
4751
Mark Scheme
June 2005
Section B
10
11
i
(x − 4)2 + 9
3
ii
(4, 9) or ft
1+1
G1
G1
iii
parabola right way up
25 at intersection on y-axis (mark
intent)
x > 7 or x < 1
iv
i
ii
iii
12
i
ii
iii
B1 for 4, B2 for 9 or M1 for 25 − 16
condone stopping at y axis
ignore posn of min: can ft theirs
2
4
M1 for x − 8x + 7 [> 0] and M1 for
(x − 7)(x − 1) [>0] or M1 for
(x − 4)2 [>] 9 and M1 for x − 4 > 3
and x − 4 < − 3 or B2 for 1 and 7
3
or [y =] (x − 4)2 − 11
1
4
1
or 1 for grad AB = 1 and grad BC =
−1 and 1 for comment/ showing
m1m2 = −1 o.e.
d or diameter needed; NB ans given
3
[y =] x2 − 8x + 5
(6 − 0)2 + (10 − 2)2
AC = 10
AB = √98 and BC = √2
clear correct use of Pythagoras’s
theorem
[angle in a semicircle so ]AC
diameter [so radius = 5]
midpt of AC = (6/2, [10+2]/2)
1
M1
A1
1
1
1
method must be shown; NB ans givn
(x − 3)2 + (y − 6)2 = 52 o.e. isw
2
B1 for one side correct
[grad AC =] 8/6 or 4/3
grad tgt = −3/4
y − 10 = [−3/4](x − 6) o.e.
[e.g. 3x + 4y = 58] or ft
(58/3, 0) and (0, 58/4) o.e. isw
(x + 1)(x − 2)(x − 5)
(x + 1)(x2 − 7x + 10)
correct step shown towards
completion [answer given]
1
M1
M1
A2
M1
A1
A1
cubic the right way up
−1, 2 and 5 indicated on x axis
10 indicated at intn on y axis
f(4) attempted
G1
G1
G1
M1
= 64 − 96+ 12 + 10
A1
M2
attempt at long division of
3
2
x − 6x + 3x + 20 by x − 4 as far as
x3 − 4x2 in working
x2 − 2x − 5 = 0
3
A2
4
for grad tgt = −1/their grad AC
or M1 for y = their m x + c then subst
(6, 10) to find c
5
1 each cao; condone not as coords
o.e. with two other factors; condone
missing brackets if expanded
correctly; A2 for x3 − 5x2 − 2x2 + x2
+ 10x − 5x − 2x + 10
must extend beyond x = −1 and 5
at intersections of curve and axis
3
3
or f(4) + 10; or ‘4 a root implies (x −
4) a factor’ or vv
or 5 × 2 × −1 etc or correct long
division if first M1 earned
or M2 for (x − 4)(x2 + .. − 5) or
(x − 4)(x2 − 2x + k) seen; M1 for
realising long divn by x − 4 needed
but not doing it
A1 for x2 − 2x − 5
SC2 for finding f(x) ÷ (x − 4) = x2 −
2x − 5 rem − 10 without further
explanation
6
Mark Scheme 4751
January 2006
Section A
1
2
3
4
5
n (n + 1) seen
= odd × even and/or even × odd
= even
M1
A1
(i) translation
⎛ 2⎞
of ⎜ ⎟
⎝0⎠
1
(ii) y = f(x − 2)
2
16 + 32x + 24x2 + 8x3 + x4 isw
4
x > −4.5 o.e. isw www
[M1 for × 4
M1 expand brackets or divide by
3
M1 subtract constant from LHS
M1 divide to find x]
[C =]
4P
−4 P
or
o.e.
1− P
P −1
1
4
4
or B1 for n odd ⇒ n2 odd, and
comment eg odd + odd = even
B1 for n even⇒ n2 even, and
comment eg even + even = even
allow A1 for ‘any number
multiplied by the consecutive
number is even’
or ‘2 to the right’ or ‘x → x + 2’
or ‘all x values are increased by
2’
1 for y = f(x + 2)
3 for 4 terms correct, 2 for 3
terms correct, or M1 for 1 4 6 4 1
s.o.i. and M1 for expansion with
correct powers of 2
accept −27/6 or better; 3 for x =
−4.5 etc
or Ms for each of the four steps
carried out correctly with
inequality [−1 if working with
equation] (ft from earlier errors if
of comparable difficulty)
M1 for PC + 4P = C
M1 for 4P = C − PC or ft
M1 for 4P = C (1 − P) or ft
B3 for
[C = ]
4
1
o.e.
−1
2
4
4
4
4
P
f(1) used
13 + 3 × 1 + k = 6
k=2
M1
A1
A1
7
grad BC = − ¼ soi
2
8
y − 3 = − ¼ (x − 2) o.e. cao
14 or ft from their BC
(i) 30√2
1
2
2
1 2
3 6
+
3 or
+
3 or
11 11
33 33
mixture of these
3
6
(ii)
unsimplified
or division by x − 1 as far as x2 +
x
or remainder = 4 + k
B3 for k = 2 www
M1 for m1m2 = −1 soi or for grad
AB = 4 or grad BC = 1/4
e.g. y = − 0.25x + 3.5
M1 for subst y = 0 in their BC
M1 for √8=2√2 or √50 = 5√2 soi
B1 for 6√50 or other correct a√b
M1 for mult num and denom by
6+√3
and M1 for denom = 11 or 33
3
5
5
3+ 6 3
1+ 2 3
or
33
11
M2 for 52 − 4k ≥ 0 or B2 for 25/4
obtained isw or M1 for b2− 4ac
soi or completing square
accept −20/8 or better, isw; M1
for attempt to express quadratic
as (2x + a)2 or for attempt at
quadratic formula
B2 for
9
(i) k ≤ 25/4
3
(ii) −2.5
2
5
Section B
10
√45 isw or 3√5
i
(0, 0),
ii
x = 3 − y or y = 3 − x seen or
used
subst in eqn of circle to
eliminate variable
9 − 6y + y2 + y2 = 45
2y2 −6y − 36 = 0 or y2 −3y − 18
=0
(y − 6)(y + 3)= 0
y = 6 or −3
x = −3 or 6
1+1
M1
M1
M1
M1
M1
A1
A1
M1
(6 − −3) 2 + (3 − −6) 2
11
i
ii
iii
iv
(x − 3.5)2 − 6.25
(3.5, −6.25) o.e. or ft from
their (i)
(0, 6) (1, 0) (6, 0)
curve of correct shape
fully correct intns and min in
4th quadrant
x2 − 7x + 6 = x2 − 3x + 4
2 = 4x
x = ½ or 0.5 or 2/4 cao
12
i
ii
iii
sketch of cubic the correct way
up
curve passing through (0, 0)
curve touching x axis at (3, 0)
x(x2 − 6x + 9) = 2
2
3
for correct expn of (3 − y)2
seen oe
condone one error if quadratic
or quad. formula attempted
[complete sq attempt earns
last 2 Ms]
8
or A1 for (6, −3) and A1 for
(−3, 6)
no ft from wrong points
(A.G.)
B1 for a = 7/2 o.e,
B2 for b = −25/4 o.e. or M1
for 6 − (7/2)2 or 6 − (their a)2
3
1+1 allow x = 3.5 and y = −6.25 or
ft; allow shown on graph
2
1 each [stated or numbers
3
shown on graph]
G1
G1
5
M1
M1 or 4x − 2 = 0 (simple linear
form; condone one error)
A1 condone no comment re only 3
one intn
G1
G1
G1
3
M1
x3 − 6x2 + 9x = 2
M1
subst x = 2 in LHS of their eqn
or in x(x − 3)2 = 2 o.e.
working to show consistent
1
division of their eqn by (x − 2)
attempted
x2 − 4x + 1
M1
1
A1
or (x2 − 3x)(x − 3) = 2 [for
one step in expanding
brackets]
for 2nd step, dep on first M1
or 2 for division of their eqn
by (x − 2) and showing no
remainder
or inspection attempted with
(x2 + kx + c) seen
2
soln of their quadratic by
formula or completing square
attempted
x = 2 ± √3 or (4 ± √12)/2 isw
locating the roots on
intersection of their curve and
y=2
M1
A2
G1
condone ignoring remainder
if they have gone wrong
A1 for one correct
must be 3 intns; condone x =
2 not marked; mark this when
marking sketch graph in (i)
7
G1
Mark Scheme 4751
June 2006
1
Section A
1
[r ] = [ ± ]
3V
o.e. ‘double-decker’
πh
3
2
a=¼
2
3
3x + 2y = 26 or y = −1.5x + 13 isw
3
4
5
6
7
8
9
3V
or r =
πh
V
1
3
πh
o.e. or M1
for correct constructive first step or for
r = k ft their r 2 = k
M1 for subst of −2 or for −8 + 4a + 7 = 0
o.e. obtained eg by division by (x + 2)
M1 for 3x + 2y = c or y = −1.5x + c
M1 for subst (2, 10) to find c or for
or for y − 10 = their gradient × (x − 2)
(i) P ⇐ Q
(ii) P ⇔ Q
x + 3(3x + 1) = 6 o.e.
1
1
M1
10x = 3 or 10y = 19 o.e.
(0.3, 1.9) or x = 0.3 and y = 1.9 o.e.
A1
A1
−3 < x < 1
[condone x < 1, x > −3]
4
(i) 28√6
2
1 for 30√6 or 2√6 or 2√2√3 or 28√2√3
(ii) 49 − 12√5 isw
3
20
2
−160 or ft for −8 × their 20
2
(i) 4/27
2
2 for 49 and 1 for − 12√5 or M1 for 3
correct terms from 4 − 6√5 − 6√5 + 45
0 for just 20 seen in second part; M1 for
6!/(3!3!) or better
condone −160x3; M1 for [−]23 × [their] 20
seen or for [their] 20 × (−2x)3; allow B1
for 160
1 for 4 or 27
(ii) 3a10b8c-2 or
10
2 for r 2 =
3a10b8
c2
M1
M1
x= ±(√10)/2 or.±√(5/2) or ±5/√10 oe
y = [±] 3√(5/2) o.e. eg y = [±] √22.5
A2
B1
2
3
condone omission of P and Q
2
for subst or for rearrangement and multn
to make one pair of coefficients the
same or for both eqns in form ‘y =’
(condone one error)
graphical soln: (must be on graph paper)
M1 for each line, A1 for (0.3, 1.9) o.e
cao; allow B3 for (0.3, 1.9) o.e.
B3 for −3 and 1 or
M1 for x2 + 2x − 3 [< 0]or (x + 1)2 < / = 4
and M1 for (x + 3)(x − 1) or x= (−2 ±4)/2
or for (x + 1) and ±2 on opp. sides of eqn
or inequality;
if 0, then SC1 for one of x < 1, x > −3
2 for 3 ‘elements’ correct, 1 for 2
elements correct, −1 for any adding of
elements; mark final answer; condone
correct but unnecessary brackets
for subst for x or y attempted
or x2 = 2.5 o.e.; condone one error from
start [allow 10x2 − 25 = 0 + correct
substn in correct formula]
allow ±√2.5; A1 for one value
ft 3 × their x value(s) if irrational;
condone not written as coords.
3
x2 + 9x2 = 25
10x2 = 25
3
2
3
4
5
4
5
5
Section B
11
i
ii
iii
12
i
ii
iii
iv
v
13
i
ii
grad AB = 8/4 or 2 or y = 2x − 10
grad BC = 1/−2 or − ½ or
y = − ½ x + 2.5
product of grads = −1 [so perp]
(allow seen or used)
1
1
midpt E of AC = (6, 4.5)
AC2 = (9 − 3)2 + (8 − 1)2 or 85
1
M1
rad = ½ √85 o.e.
(x − 6)2 + (y − 4.5)2 = 85/4 o.e.
A1
B2
(5−6)2 + (0−4.5)2 = 1 + 81/4 [=
85/4]
1
uuur uuur ⎛ 1 ⎞
BE = ED = ⎜ ⎟
⎝ 4.5 ⎠
iv
allow seen in (i) only if used in (ii); or
AE2 = (9 − their 6)2 + (8 − their 4.5)2 or
rad.2 = 85/4 o.e. e.g. in circle eqn
M1 for (x − a)2 + (y − b)2 = r2 soi or for
lhs correct
some working shown; or ‘angle in
semicircle [=90°]’
uuur
o.e. ft their centre; or for BC = ⎜
M1
or (9 − 2, 8 + 1); condone mixtures of
vectors and coords. throughout part iii
allow B3 for (7,9)
M1
A1
divn attempted as far as x2 + 3x
x2 + 3x + 2 or (x + 2)(x + 1)
(x + 2)(x + 6)(x + 1)
M1
A1
2
A1
or M1 for division by (x + 2) attempted
as far as x3 + 2x2 then A1 for x2 + 7x +
6 with no remainder
or inspection with b = 3 or c = 2 found;
B2 for correct answer
allow seen earlier;
M1 for (x + 2)(x + 1)
with 2 turning pts; no 3rd tp
curve must extend to x > 0
condone no graph for x < −6
or other partial factorisation
sketch of cubic the right way up
through 12 marked on y axis
intercepts −6, −2, −1 on x axis
[x](x2 + 9x + 20)
[x](x + 4)(x + 5)
x = 0, −4, −5
G1
G1
G1
M1
M1
A1
y = 2x + 3 drawn on graph
x = 0.2 to 0.4 and − 1.7 to −1.9
M1
A2
1 = 2x2 + 3x
2x2 + 3x − 1 [= 0]
M1
M1
1 each; condone coords; must have
line drawn
for multiplying by x correctly
for correctly rearranging to zero (may
be earned first) or suitable step re
completing square if they go on
attempt at formula or completing
square
M1
ft, but no ft for factorising
A2
A1 for one soln
1
and approaching y = 2 from above
1
2
and extending below x axis
1 each; may be found algebraically;
ignore y coords.
−3 ± 17
4
branch through (1,3),
branch through (−1,1),approaching
y = 2 from below
−1 and ½ or ft intersection of their
curve and line [tolerance 1 mm]
3
6
⎛ −2 ⎞
⎟
⎝1⎠
M1
D has coords (6 + 1, 4.5 + 4.5) ft
or
(5 + 2, 0 + 9)
= (7, 9)
f(−2) used
−8 + 36 − 40 + 12 = 0
x=
iii
1
or M1 for AB2 = 42 + 82 or 80 and
BC2 = 22 + 12 or 5 and AC2 = 62 + 72 or
85; M1 for AC2 = AB2 + BC2 and 1 for
[Pythag.] true so AB perp to BC;
3
if 0, allow G1 for graph of A, B, C
or B1 for each root found e.g. using
factor theorem
3
2
2
2
3
3
3
5
2
2
Mark Scheme 4751
January 2007
1
4751
Mark Scheme
Jan 2007
Section A
1
2
3
y = 2x + 4
neg quadratic curve
intercept (0, 9)
through (3, 0) and (−3, 0)
[ a =]
2c
− 2c
or
as final answer
2− f
f −2
3
1
1
1
3
M1 for m = 2 stated [M0 if go on to use
m = − ½ ] or M1 for y = 2x + k, k ≠ 7
and M1indep for y − 10 = m(x − 3) or (3,
10) subst in y = mx + c; allow 3 for y = 2x
+ k and k = 4
condone (0, 9) seen eg in table
3
M1 for attempt to collect as and cs on
different sides and M1 ft for a (2 − f) or
dividing by 2 − f; allow M2 for
7c − 5c
2− f
etc
4
5
f(2) = 3 seen or used
M1
23 + 2k + 5 = 3 o.e.
k = −5
M1
B1
375
3
allow 375x4; M1 for 52 or 25 used or
seen with x4 and
6×5
6!
oe eg
or 1 6 15
2
4!2!
… seen [6C4 not sufft]
(i) 125
2
M1 for 25 2 = 25 soi or for
9
as final answer
49
2
M1 for a −1 =
showing a + b + c = 6 o.e
1
simple equiv fraction eg 192/32 or 24/4
M1
correct expansion of numerator; may be
unsimplified 4 term expansion; M0 if get
(ii)
7
bc =
9 − 17
16
2
3
allow M1 for divn by (x − 2) with x2 + 2x +
(k + 4) or x2 + 2x − 1 obtained
alt: M1 for (x − 2)(x2 + 2x − 1) + 3 (may
be seen in division) then M1dep (and
B1) for x3 − 5x + 5
alt divn of x3 + kx + 2 by x − 2 with no
3
rem.
M1 for 15 or
6
3
1
253
1
soi eg by 3/7 or 3/49
a
no further than
(
)
3
4
2
17 ; M0 if no
evidence before 64/16 o.e.
=64/16 o.e. correctly obtained
A1
completion showing abc = 6 o.e.
A1
may be implicit in use of factors in
completion
4
2
4751
8
Mark Scheme
b 2 − 4ac soi
use of b 2 − 4ac < 0
k2 < 16 [may be implied by k < 4]
−4 < k < 4 or k > −4 and k < 4 isw
9
(i) 12a5b3 as final answer
( x + 2)( x − 2)
( x − 2)( x − 3)
x+2
as final answer
x−3
(ii)
10
11
M1
M1
A1
A1
1 for 2 ‘terms’ correct in final answer
M2
M1 for each of numerator or denom.
correct or M1, M1 for correct factors
seen separately
4m2 correctly obtained
[p =] [±]2m cao
A1
A1
Section B
iA
0.2 to 0.3 and 3.7 to 3.8
1
= 4− x
x
their y = 4 − x drawn
0.2 to 0.35 and 1.65 to 1.8
ii
iii
(0, ±√3)
centre (1, 0) radius 2
touches at (1, 2) [which is distance
2 from centre]
all points on other branch > 2 from
centre
4
5
A1
M2
x+
may be implied by k2 < 16
deduct one mark in qn for ≤ instead of <;
allow equalities earlier if final inequalities
correct; condone b instead of k; if M2 not
earned, give SC2 for qn [or M1 SC1] for
k [=] 4 and − 4 as answer]
2
correct expansion of both brackets
seen (may be unsimplified), or
difference of squares used
iB
January 2007
M1 for one bracket expanded correctly;
for M2, condone done together and lack
of brackets round second expression if
correct when we insert the pair of
brackets
4
1+1
[tol. 1mm or 0.05 throughout qn]; if 0,
allow M1 for drawing down lines at
both values
2
M1
condone one error
M1
allow M2 for plotting positive branch of
y = 2x + 1/x [plots at (1,3) and (2,4.5)
and above other graph] or for plot of y
= 2x2 − 4x + 1
B2
1 each
4
2
condone y = ±√3 isw; 1 each or
M1 for 1 + y2 = 4 or y2 = 3 o.e.
2
allow seen in (ii)
1+1
1
allow ft for both these marks for centre
at (−1, 0), rad 2;
allow 2 for good sketch or compassdrawn circle of rad 2 centre (±1, 0)
1
3
4
4751
12
i
ii
Mark Scheme
(3, 6)
2
1 each coord
grad AB = (8 − 4)/(7 − −1) or 4/8
grad normal = −2 or ft
M1
M1
perp bisector is
y − 6 = −2(x − 3) or ft their grad. of
normal (not AB) and/or midpoint
correct step towards completion
indep obtained
for use of m1m2 = −1; condone
stated/used as −2 with no working
only if 4/8 seen
M1
A1
or M1 for showing grad given line = −2
and M1 for showing (3, 6) fits given
line
Bisector crosses y axis at C (0, 12)
seen or used
AB crosses y axis at D (0, 4.5)
seen or used
M1
may be implicit in their area calcn
B2
½ × (12 − their 4.5) × 3
(may be two triangles M1 each)
M2
M1 for 4 + their grad AB or for eqn AB
is y − 8 = their ½ (x − 7) oe with
coords of A or their M used
or M1 for [MC]2 = 32 + 62 or 45 or
[MD]2 = 32 + 1.52 or 11.25 oe and M1
for ½ × their MC × MD; all ft their M
45/4 o.e. without surds, isw
y
A1
D
A
(−1, 4)
M
B
(7, 8)
0
MR: intn used as D(0, 4) can score a
max of M1, B0, M2 (eg M1 for their
DM = √13), A0
x
condone poor notation
alt allow integration used:
∫
3
0
(−2 x + 12)dx [= 27]
M1
allow if seen, with correct line and
limits seen/used
as above
M1
obtaining AB is y − 8 = their ½ (x −
7) oe [y = ½ x + 4.5]
∫
3
0
( 12 x + 4.5)dx
= 63/4 o.e. cao
their area under CB − their area
under AB
M1
ft from their AB
A1
M1
M1
M1
allow only if at least some valid
integration/area calculations for these
trapezia seen
if combined integration, so 63/4 not
found separately, mark equivalently
for Ms and allow A2 for final answer
eg may be implied by divn
or other factor (x2 ….−1) or (x2 + 2x...)
A1
M1
or B3 www
ft their quadratic
A1
= 45/4 o.e. cao
i
6
MR: AMC used not DMC: lose B2 for
D but then allow ft M1 for MC2 or MA2
[=42 + 22] and M1 for ½ × MA × MC
and A1 for 15
C
13
January 2007
x − 2 is factor soi
attempt at divn by x − 2 as far as
x3 − 2x2 seen in working
x2 + 2x − 1 obtained
attempt at quad formula or comp
square
−1 ± 2 as final answer
A2
A1 for
4
−2 ± 8
seen; or B3 www
2
6
6
4751
ii
iii
Mark Scheme
f(x − 3) = (x − 3)3 − 5(x − 3) + 2
B1
(x − 3)(x2 − 6x + 9) or other
constructive attempt at expanding
(x − 3)3 eg 1 3 3 1 soi
M1
x3 − 9x2 + 27x − 27
− 5x + 15 [+2]
A1
B1
5
B1
B1
2 ± 2 or ft
January 2007
or (x − 5)(x − 2 + 2 )(x − 2 − 2 ) soi
or ft from their (i)
for attempt at multiplying out 2
brackets or valid attempt at multiplying
all 3
5
alt: A2 for correct full unsimplified
expansion or A1 for correct 2 bracket
expansion eg (x − 5)(x2 − 4x + 2)
4
condone factors here, not roots
if B0 in this part, allow SC1 for their
roots in (i) − 3
2
4751/01
ADVANCED SUBSIDIARY GCE UNIT
MATHEMATICS (MEI)
Introduction to Advanced Mathematics (C1)
INSERT
TUESDAY 16 JANUARY 2007
Morning
Time: 1 hour 30 minutes
Candidate
Name
Candidate
Number
Centre
Number
INSTRUCTIONS TO CANDIDATES
•
•
This insert should be used in Question 11.
Write your name, centre number and candidate number in the spaces provided above and attach the
page to your answer booklet.
This insert consists of 2 printed pages.
HN/2
© OCR 2007 [L/102/2657]
OCR is an exempt Charity
[Turn over
2
11 (i)
y
8
7
y=x+
6
1
x
5
4
3
2
1
–6
–5
–4
–3
–2
–1 0
1
–1
–2
–3
–4
–5
–6
–7
–8
–9
–10
© OCR 2007
4751/01 Insert Jan 07
2
3
4
5
6
x
Mark Scheme 4751
January 2007
1
4751
Mark Scheme
Jan 2007
Section A
1
2
3
y = 2x + 4
neg quadratic curve
intercept (0, 9)
through (3, 0) and (−3, 0)
[ a =]
2c
− 2c
or
as final answer
2− f
f −2
3
1
1
1
3
M1 for m = 2 stated [M0 if go on to use
m = − ½ ] or M1 for y = 2x + k, k ≠ 7
and M1indep for y − 10 = m(x − 3) or (3,
10) subst in y = mx + c; allow 3 for y = 2x
+ k and k = 4
condone (0, 9) seen eg in table
3
M1 for attempt to collect as and cs on
different sides and M1 ft for a (2 − f) or
dividing by 2 − f; allow M2 for
7c − 5c
2− f
etc
4
5
f(2) = 3 seen or used
M1
23 + 2k + 5 = 3 o.e.
k = −5
M1
B1
375
3
allow 375x4; M1 for 52 or 25 used or
seen with x4 and
6×5
6!
oe eg
or 1 6 15
2
4!2!
… seen [6C4 not sufft]
(i) 125
2
M1 for 25 2 = 25 soi or for
9
as final answer
49
2
M1 for a −1 =
showing a + b + c = 6 o.e
1
simple equiv fraction eg 192/32 or 24/4
M1
correct expansion of numerator; may be
unsimplified 4 term expansion; M0 if get
(ii)
7
bc =
9 − 17
16
2
3
allow M1 for divn by (x − 2) with x2 + 2x +
(k + 4) or x2 + 2x − 1 obtained
alt: M1 for (x − 2)(x2 + 2x − 1) + 3 (may
be seen in division) then M1dep (and
B1) for x3 − 5x + 5
alt divn of x3 + kx + 2 by x − 2 with no
3
rem.
M1 for 15 or
6
3
1
253
1
soi eg by 3/7 or 3/49
a
no further than
(
)
3
4
2
17 ; M0 if no
evidence before 64/16 o.e.
=64/16 o.e. correctly obtained
A1
completion showing abc = 6 o.e.
A1
may be implicit in use of factors in
completion
4
2
4751
Mark Scheme
4751 (C1)
January 2008
Introduction to Advanced Mathematics
Section A
1
[v = ] [ ± ]
2E
www
m
3
M2 for v 2 =
2E
or for [v =] [ ± ]
m
E
or
m
1
2
M1 for a correct constructive first step
and M1 for v = [ ± ] k ft their v2 = k;
2
3x − 4
7
or 3 −
www as final
x +1
x +1
if M0 then SC1 for √E/ ½ m or √2E/m
etc
M1 for (3x − 4)(x − 1)
and M1 for (x + 1)(x − 1)
3
3
answer
3
4
5
3
(i) 1
1
(ii) 1/64 www
3
6x + 2(2x − 5) = 7
M1
10x = 17
M1
x = 1.7 o.e. isw
y = −1.6 o.e .isw
A1
A1
M1 for dealing correctly with each of
reciprocal, square root and cubing
(allow 3 only for 1/64)
eg M2 for 64 or −64 or 1/√4096 or ¼3
or M1 for 1/163/2 or 43 or −43 or 4-3 etc
4
for subst or multn of eqns so one pair of
coeffts equal (condone one error)
simplification (condone one error) or
appropriate addn/subtn to eliminate
variable
allow as separate or coordinates as
requested
graphical soln: M0
4
(i) −4/5 or −0.8 o.e.
2
M1 for 4/5 or 4/−5 or 0.8 or −4.8/6 or
correct method using two points on the
line (at least one correct) (may be
graphical) or for −0.8x o.e.
(ii) (15, 0) or 15 found www
3
M1 for y = their (i) x + 12 o.e. or 4x + 5y
= k and (0, 12) subst and M1 for using y
= 0 eg −12 = −0.8x or ft their eqn
or M1 for given line goes through (0,
4.8) and (6, 0) and M1 for 6 × 12/4.8
graphical soln: allow M1 for correct
required line drawn and M1 for answer
within 2mm of (15, 0)
1
5
19
4751
6
Mark Scheme
f(2) used
M1
23 + 2k + 7 = 3
M1
k = −6
7
8
9
January 2008
or division by x − 2 as far as x2 + 2x
obtained correctly
or remainder 3 = 2(4 + k) + 7 o.e. 2nd
M1 dep on first
A1
(i) 56
2
(ii) −7 or ft from −their (i)/8
2
3
M1 for
8× 7 × 6
or more simplified
3 × 2 ×1
M1 for 7 or ft their (i)/8 or for
56 × (−1/2)3 o.e. or ft; condone x3 in
answer or in M1 expression;
0 in qn for just Pascal’s triangle seen
(i) 5√3
2
M1 for √48 = 4√3
(ii) common denominator =
(5 − √2)(5 + √2)
=23
numerator = 10
M1
A1
B1
allow M1A1 for
5− 2 5+ 2
+
23
23
allow 3 only for 10/23
(i) n = 2m
M1
or any attempt at generalising; M0 for
just trying numbers
3n2 + 6n = 12m2 + 12m or
= 12m(m + 1)
M2
or M1 for 3n2 + 6n = 3n (n + 2) = 3 ×
even × even and M1 for explaining that
4 is a factor of even × even
or M1 for 12 is a factor of 6n when n is
even and M1 for 4 is a factor of n2 so 12
is a factor of 3n2
(ii) showing false when n is odd e.g.
3n2 + 6n = odd + even = odd
B2
or 3n (n + 2) =3 × odd × odd = odd or
counterexample showing not always
true; M1 for false with partial
explanation or incorrect calculation
2
4
5
5
17
4751
Mark Scheme
January 2008
Section B
10
i
ii
iii
correct graph with clear
asymptote x = 2 (though need not
be marked)
(0, − ½ ) shown
11/5 or 2.2 o.e. isw
x=
1
x−2
x(x − 2) = 1 o.e.
x2 − 2x − 1 [= 0]; ft their equiv
eqn
attempt at quadratic formula
1 ±√2 cao
position of points shown
11
i
(x − 2.5)2 o.e.
− 2.52 + 8
(x − 2.5)2 + 7/4 o.e.
min y = 7/4 o.e. [so above x axis]
or commenting (x − 2.5)2 ≥ 0
ii
iii
iv
G2
G1 for one branch correct; condone
(0, − ½ ) not shown
SC1 for both sections of graph
shifted two to left
G1
allow seen calculated
3
2
M1 for correct first step
2
M1
or equivs with ys
M1
M1
M1
A1
B1
M1
M1
A1
B1
or (x − 1)2 − 1 = 1 o.e.
or (x − 1) = ±√2 (condone one error)
on their curve with y = x (line drawn
or y = x indicated by both coords);
condone intent of diagonal line with
gradient approx 1through origin as y
= x if unlabelled
11
for clear attempt at −2.52
allow M2A0 for (x − 2.5) + 7/4 o.e.
with no (x − 2.5)2 seen
ft, dep on (x − a)2 + b with b positive;
condone starting again, showing b2 −
4ac < 0 or using calculus
correct symmetrical quadratic
shape
8 marked as intercept on y axis
tp (5/2, 7/4) o.e. or ft from (i)
G1
G1
G1
or (0, 8) seen in table
x2 − 5x − 6 seen or used
−1 and 6 obtained
x < − 1 and x > 6 isw or ft their
solns
M1
M1
M1
or (x − 2.5)2 [> or =] 12.25 or ft 14 − b
also implies first M1
if M0, allow B1 for one of x < − 1 and
x>6
min = (2.5, − 8.25) or ft from (i)
so yes, crosses
M1
A1
3
6
4
3
or M1 for other clear comment re
translated 10 down and A1 for
referring to min in (i) or graph in (ii);
or M1 for correct method for solving
x2 −5x −2 = 0 or using b2 − 4ac with
this and A1 for showing real solns eg
b2 − 4ac = 33; allow M1A0 for valid
comment but error in −8.25 ft; allow
M1 for showing y can be neg eg (0,
−2) found and A1 for correct
conclusion
3
2
12
4751
12
Mark Scheme
i
(x − 4)2 − 16 + (y − 2)2 − 4 = 9
o.e.
M2
B1
rad = √29
ii
iii
iv
42 + 22 o.e
= 20 which is less than 29
M1
A1
showing midpt of AB = (4, 2)
and
showing AB = 2√29 or showing
AC or BC = √29 or that A or B lie
on circle
2
or showing both A and B lie on
circle (or AC = BC = √29),
and
showing AB = 2√29 or that C is
midpt of AB or that C is on AB
or that gradients of AB and AC
are the same or equiv.
2
or showing C is on AB
and
showing both A and B are on
circle or AC = BC = √29
2
grad AC or AB or BC = −5/2 o.e.
M1
grad tgt = −1/their grad AC
tgt is y − 7 = their m (x − 2) o.e.
M1
M1
y = 2/5 x + 31/5 o.e.
A1
2
2
January 2008
M1 for one completing square or for
(x − 4)2 or (y − 2)2 expanded correctly
or starting with (x − 4)2 + (y − 2)2 = r2:
M1 for correct expn of at least one
bracket and M1 for 9 + 20 = r2 o.e.
or using x2 −2gx + y2 − 2fy+ c = 0
M1 for using centre is (g, f) [must be
quoted] and M1 for r2 = g2 + f2 − c
3
allow 2 for showing circle crosses x
axis at −1 and 9 or equiv for y (or
showing one positive; one negative);
0 for graphical solutions (often using
A and B from (iii) to draw circle)
2
in each method, two things need to
be established. Allow M1 for the
concept of what should be shown
and A1 for correct completion with
method shown
allow M1A0 for AB just shown as
√116 not 2√29
allow M1A0 for stating mid point of
AB = (4,2) without working/method
shown
NB showing AB = 2√29 and C lies on
AB is not sufficient – earns 2 marks
only
2
4
if M0, allow SC2 for accurate graph
of circle drawn with compasses and
AB joined with ruled line through C.
may be seen in (iii) but only allow this
M1 if they go on to use in this part
allow for m1m2=−1 used
eg y = their mx + c then (2, 7) subst;
M0 if grad AC used
condone y = 2/5x + c and c = 31/5
o.e.
4
4
13
4751
Mark Scheme
June 2008
4751 (C1) Introduction to Advanced Mathematics
Section A
x > 6/4 o.e. isw
1
2
3
4
5
2
(i) (0, 4) and (6, 0)
2
(ii) −4/6 o.e. or ft their (i) isw
2
M1 for 4x > 6 or for 6/4 o.e. found or for
their final ans ft their 4x > k or kx > 6
2
1 each; allow x = 0, y = 4 etc; condone
x = 6, y = 4 isw but 0 for (6, 4) with no
working
1 for − 64 x or 4/−6 or 4/6 o.e. or ft
(accept 0.67 or better)
0 for just rearranging to y = − 23 x + 4
4
(i) 0 or −3/2 o.e.
2
1 each
(ii) k < −9/8 o.e. www
3
M2 for 32 (−)(−8k) < 0 o.e. or −9/8 found
or M1 for attempted use of
b2 − 4ac (may be in quadratic formula);
SC: allow M1 for 9 − 8k < 0 and M1 ft
for k > 9/8
3 for all correct, 2 for 3 correct. 1 for 2
correct
(i) T
(ii) E
(iii) T
(iv) F
3
y (x − 2) = (x + 3)
M1
for multiplying by x − 2; condone
missing brackets
xy − 2y = x + 3 or ft [ft from earlier
errors if of comparable difficulty – no
ft if there are no xy terms]
M1
for expanding bracket and being at
stage ready to collect x terms
xy − x = 2y + 3 or ft
M1
for collecting x and ‘other’ terms on
opposite sides of eqn
M1
for factorising and division
[ x =]
5
3
2y + 3
o.e. or ft
y −1
for either method: award 4 marks only if
fully correct
alt method:
y = 1+
5
x−2
5
x−2
5
x−2 =
y −1
y −1 =
5
x = 2+
y −1
M1
M1
M1
M1
4
1
18
4751
6
7
Mark Scheme
(i) 5 www
2
allow 2 for ±5; M1 for 251/2 seen or for
1/5 seen or for using 251/2 = 5 with
another error (ie M1 for coping correctly
with fraction and negative index or with
square root)
(ii) 8x10y13z4 or 23x10y13z4
3
mark final answer; B2 for 3 elements
correct, B1 for 2 elements correct;
condone multn signs included, but −1
from total earned if addn signs
(i)
5− 3
5 + (−1) 3
5 −1 3
or
or
22
22
22
(ii) 37 − 12√ 7 isw www
8
June 2008
−2000 www
2
5
or a = 5, b = −1, c = 22; M1 for attempt
to multiply numerator and denominator
by 5 − 3
2 for 37 and 1 for − 12√ 7 or M1 for 3
correct terms from 9 − 6√ 7 − 6√ 7 + 28
or 9 − 3√ 28 − 3√ 28 + 28 or 9 − √ 252 −
√ 252 + 28 o.e. eg using 2√ 63
or M2 for 9 − 12√ 7 + 28 or 9 − 6√ 28 +
28 or 9 − 2√ 252 + 28 or 9 − √ 1008 +
28 o.e.; 3 for 37 − √ 1008 but not other
equivs
3
5
M3 for 10 × 52 × (−2[x])3 o.e. or M2 for
two of these elements or M1 for 10 or
(5×4×3)/(3×2×1) o.e. used [5C3 is not
sufficient] or for 1 5 10 10 5 1 seen;
4
or B3 for 2000;
condone x3 in ans;
⎛ 2 ⎞
[ x] ⎟
⎝ 5 ⎠
3
equivs: M3 for e.g 55 ×10 × ⎜ −
o.e. [55 may be outside a bracket for
whole expansion of all terms], M2 for
two of these elements etc
similarly for factor of 2 taken out at start
9
(y − 3)(y − 4) [= 0]
M1
y = 3 or 4 cao
A1
x = ± 3 or ± 2 cao
B2
4
for factors giving two terms correct or
attempt at quadratic formula or
completing square
or B2 (both roots needed)
B1 for 2 roots correct or ft their y
(condone √ 3 and √ 4 for B1)
2
4
18
4751
Mark Scheme
Section B
10 i
(x − 3)2 − 7
3
June 2008
mark final answer; 1 for a = 3,
2 for b = 7 or M1 for −32 + 2;
bod 3 for (x − 3) − 7
ii
(3, −7) or ft from (i)
1+1
iii
sketch of quadratic correct way
up and through (0, 2)
G1
accept (0, 2) o.e. seen in this part [eg
in table] if 2 not marked as intercept
on graph
t.p. correct or ft from (ii)
G1
accept 3 and −7 marked on axes
level with turning pt., or better; no ft
for (0, 2) as min
iv
3
2
x2 − 6x + 2 = 2x − 14 o.e.
M1
or their (i) = 2x − 14
x2 − 8x + 16 [= 0]
M1
dep on first M1; condone one error
(x − 4)2 [= 0]
M1
or correct use of formula, giving
equal roots; allow (x + 4)2 o.e.
ft x2 + 8x + 16
x = 4, y = −6
A1
if M0M0M0, allow SC2 for showing
(4, −6) is on both graphs (need to go
on to show line is tgt to earn more)
equal/repeated roots [implies tgt] must be explicitly stated; condone
‘only one root [so tgt]’ or ‘line
meets curve only once, so tgt’ or
‘line touches curve only once’ etc]
A1
or for use of calculus to show grad of
line and curve are same when x = 4
2
5
3
12
4751
11
Mark Scheme
i
ii
iii
f(−4) used
M1
−128 + 112 + 28 − 12 [= 0]
A1
June 2008
or B2 for (x + 4)(2x2 − x − 3) here; or
correct division with no remainder
division of f(x) by (x + 4)
M1
as far as 2x3 + 8x2 in working, or two
terms of 2x2 − x − 3 obtained by
inspection etc (may be earned in (i)),
or f(−1) = 0 found
2x2 − x − 3
A1
2x2 − x − 3 seen implies M1A1
(x + 1)(2x − 3)
A1
[f(x) =] (x + 4) (x + 1)(2x − 3)
A1
or B4; allow final A1 ft their factors if
M1A1A0 earned
sketch of cubic correct way up
G1
ignore any graph of y = f(x − 4)
through −12 shown on y axis
G1
or coords stated near graph
roots −4, −1, 1.5 or ft shown on x
axis
G1
or coords stated near graph
if no curve drawn, but intercepts
marked on axes, can earn max of
G0G1G1
iv
x (x − 3)(2[x − 4] −3) o.e. or
x (x − 3)(x − 5.5) or ft their factors
M1
or
2(x − 4)3 + 7(x − 4)2 − 7(x − 4) − 12
or stating roots are 0, 3 and 5.5 or ft;
condone one error
eg 2x − 7 not 2x − 11
correct expansion of one pair of
brackets ft from their factors
M1
or for correct expn of (x − 4)3 [allow
unsimplified]; or for showing g(0) =
g(3) = g(5.5) = 0 in given ans g(x)
correct completion to given
answer
M1
allow M2 for working backwards from
given answer to x(x − 3)(2x − 11) and
M1 for full completion with factors or
roots
2
4
3
3
4
12
4751
12
Mark Scheme
i
grad AB =
9 −1
or 2
3 − −1
y − 9 = 2(x − 3) or y − 1 = 2(x + 1)
ii
M1
M1
A1
y = 2x + 3 o.e.
M1
grad perp = −1/grad AB
M1
y − 5 = − ½ (x − 1) o.e. or ft [no ft
for just grad AB used]
M1
ft their grad and/or midpt, but M0
if their midpt not used; allow M1
for y = − ½x + c and then their
midpt subst
at least one correct interim step
towards given answer 2y + x =
11, and correct completion
NB ans 2y + x = 11 given
M1
no ft; correct eqn only
y=
11 − x
o.e.
2
condone not stated explicitly, but
used in eqn
soi by use eg in eqn
M1
grad perp = −1/grad AB and
showing/stating same as given
line
M1
eg stating − ½ × 2 = −1
finding intn of their y = 2x + 3
M1
or showing that (1, 5) is on 2y + x
= 11, having found (1, 5) first
showing midpt of AB is (1, 5)
M1
[for both methods: for M4 must be
fully correct]
showing (−1 − 5)2 + (1 − 3)2 = 40
M1
showing B to centre = √40 or
verifying that (3, 9) fits given circle
M1
and 2y + x = 11 [= (1, 5)]
(x − 5)2 + 32 = 40
M1
(x − 5)2 = 31
M1
x = 5 ± 31 or
3
mark one method or the other, to
benefit of cand, not a mixture
ans:
iv
ft their m, or subst coords of A or B in
y = their m x + c
or B3
mid pt of AB = (1, 5)
alt method working back from
iii
June 2008
10 ± 124
isw
2
A1
5
at least one interim step needed for
each mark; M0 for just 62 + 22 = 40
with no other evidence such as a first
line of working or a diagram;
condone marks earned in reverse
order
for subst y = 0 in circle eqn
4
2
condone slip on rhs; or for
rearrangement to zero (condone one
error) and attempt at quad. formula
[allow M1 M0 for (x − 5)2 = 40 or for
(x − 5)2 +32 = 0 ]
or 5 ±
124
2
3
12
4751
Mark Scheme
January 2009
4751 (C1) Introduction to Advanced Mathematics
Section A
(i) 0.125 or 1/8
1
(ii) 1
2
3
4
5
y = 5x − 4 www
x > 9/6 o.e. or 9/6 < x o.e. www isw
a = −5 www
1
1
as final answer
3
y − 11
x −3
o.e.
=
−9 − 11 −1 − 3
11 − ( −9 )
or 5 eg in y
or M1 for grad =
3 − ( −1)
= 5x + k and M1 for y − 11 = their m(x −
3) o.e. or subst (3, 11) or (−1, −9) in
y = their mx + c or M1 for y = kx − 4 (eg
may be found by drawing)
3
M2 for 9 < 6x or M1 for −6x < −9 or k <
6x or 9 < kx or 7 + 2 < 5x + x [condone
≤ for Ms];
if 0, allow SC1 for 9/6 o.e found
3
M1 for f(2) = 0 used and M1 for 10 +
2a = 0 or better
long division used:
M1 for reaching (8 + a)x − 6 in working
and M1 for 8 + a = 3
equating coeffts method:
M2 for obtaining x3 + 2x2 + 4x + 3 as
other factor
3
2
M2 for
3
3
(i) 4[x3]
2
ignore any other terms in expansion
M1 for −3[x3] and 7[x3] soi;
(ii) 84[x2] www
3
7×6
or 21 or for Pascal’s
2
triangle seen with 1 7 21 … row
and M1 for 22 or 4 or {2x}2
M1 for
1
5
16
4751
6
7
Mark Scheme
1/5 or 0.2 o.e. www
(i) 53.5 or k = 3.5 or 7/2 o.e.
January 2009
M1 for 3x + 1 = 2x × 4 and
M1 for 5x = 1 o.e.
or
1
= 4 and
M1 for 1.5 +
2x
1
M1 for
= 2.5 o.e.
2x
3
2
3
M1 for 125 = 5 or
5 =5
3
1
2
3
2
SC1 for 5 o.e. as answer without
working
(ii) 16a6b10
8
9
10
2
M1 for two ‘terms’ correct and
multiplied; mark final answer only
4
b2 − 4ac soi
M1
allow in quadratic formula or clearly
looking for perfect square
k2 − 4 × 2 × 18 < 0 o.e.
M1
−12 < k < 12
A2
y + 5 = xy + 2x
y − xy = 2x − 5 oe or ft
y (1 − x) = 2x − 5 oe or ft
2x − 5
[ y =]
oe or ft as final answer
1− x
M1
M1
M1
M1
condone ≤; or M1 for 12 identified as
boundary
may be two separate inequalities; A1 for
≤ used or for one ‘end’ correct
if two separate correct inequalities seen,
isw for then wrongly combining them
into one statement;
condone b instead of k;
4
if no working, SC2 for k < 12 and SC2
for k > −12 (ie SC2 for each ‘end’
correct)
for expansion
for collecting terms
for taking out y factor; dep on xy term
for division and no wrong work after
ft earlier errors for equivalent steps if
error does not simplify problem
(i) 9 3
2
M1 for 5√3 or 4√3 seen
(ii) 6 + 2√2 www
3
M1 for attempt to multiply num. and
denom. by 3 + √2 and M1 for denom. 7
or 9 − 2 soi from denom. mult by 3 + √2
2
4
5
20
4751
Mark Scheme
Section B
11 i
ii
iii
January 2009
⎛ 11 + ( −1) 4 ⎞
C, mid pt of AB = ⎜
, ⎟
2
2⎠
⎝
= (5, 2)
B1
evidence of method required – may
be on diagram, showing equal steps,
or start at A or B and go half the
difference towards the other
[AB2 =] 122 + 42 [= 160] oe or
[CB2 =] 62 + 22 [=40] oe with AC
B1
or square root of these; accept
unsimplified
quote of (x − a)2 + (y − b)2 = r2
o.e
with different letters
B1
or (5, 2) clearly identified as centre
and 40 as r (or 40 as r2) www
or quote of gfc formula and finding c
= −11
completion (ans given)
B1
dependent on centre (or midpt) and
radius (or radius2) found
independently and correctly
4
condone one error
or use of quad formula (condone one
error in formula); ft only for 3 term
quadratic in y
if y = 0 subst, allow SC1 for (11, 0)
found
alt method:
M1 for y values are 2 ± a
M1 for a2 + 52 = 40 soi
M1 for a2 = 40 − 52 soi
A1 for [ y =]2 ± 15 cao
4
correct subst of x = 0 in circle eqn
soi
(y − 2)2 = 15 or y2 − 4y − 11 [= 0]
y − 2 = ± 15 or ft
M1
[ y =]2 ± 15 cao
A1
M1
M1
4
or 1/3 o.e.
11 − ( −1)
M1
or grad AC (or BC)
so grad tgt = − 3
eqn of tgt is y − 4 = − 3 (x − 11)
M1
M1
y = −3x + 37 or 3x + y = 37
(0, 37) and (37/3, 0) o.e. ft isw
A1
B2
or ft −1/their gradient of AB
or subst (11, 4) in y = − 3x + c or ft
(no ft for their grad AB used)
accept other simplified versions
B1 each, ft their tgt for grad ≠ 1 or
1/3; accept x = 0, y = 37 etc
NB alt method: intercepts may be
found first by proportion then used to
find eqn
grad AB =
3
6
14
4751
12
Mark Scheme
i
ii
iii
3x2 + 6x + 10 = 2 − 4x
M1
3x2 + 10x + 8 [= 0]
M1
(3x + 4)(x + 2) [=0]
M1
x = −2 or −4/3 o.e.
y = 10 or 22/3 o.e.
A1
A1
3(x + 1)2 + 7
4
min at y = 7 or ft from (ii) for
positive c (ft for (ii) only if in
correct form)
B2
4
January 2009
for subst for x or y or subtraction
attempted
or 3y2 − 52y + 220 [=0]; for
rearranging to zero (condone one
error)
or (3y − 22)(y − 10); for sensible
attempt at factorising or formula or
completing square
or A1 for each of (−2, 10) and
(−4/3, 22/3) o.e.
5
1 for a = 3, 1 for b = 1, 2 for c = 7 or
M1 for 10 − 3 × their b2 soi or for 7/3
or for 10/3 − their b2 soi
4
may be obtained from (ii) or from
good symmetrical graph or identified
from table of values showing
symmetry
condone error in x value in stated min
ft from (iii) [getting confused with 3
factor]
B1 if say turning pt at y = 7 or ft
without identifying min
or M1 for min at x = −1 [e.g. may
start again and use calculus to obtain
x = −1] or min when (x + 1)[2] = 0;
and A1 for showing y positive at min
or M1 for showing discriminant neg.
so no real roots and A1 for showing
above axis not below eg positive x2
term or goes though (0, 10)
or M1 for stating bracket squared
must be positive [or zero] and A1 for
saying other term is positive
2
11
4751
13
Mark Scheme
i
January 2009
any correct y value calculated
from quadratic seen or implied by
plots
B1
for x ≠ 0 or 1; may be for neg x or eg
min.at (2.5, −1.25)
(0, 5)(1, 1)(2, −1)(3, −1)(4,1) and
(5,5) plotted
P2
tol 1 mm; P1 for 4 correct [including
(2.5, −1.25) if plotted]; plots may be
implied by curve within 1 mm of
correct position
good quality smooth parabola
within 1mm of their points
C1
allow for correct points only
[accept graph on graph paper, not
insert]
ii
iii
1
x
x3 − 5x2 + 5x = 1 and completion
to given answer
x2 − 5x + 5 =
4
M1
M1
2
divn of x3 − 5x2 + 5x − 1 by x − 1
as far as x3 − x2 used in working
M1
x2 − 4x + 1 obtained
A1
use of b 2 − 4ac or formula with
quadratic factor
12 obtained and comment re
shows other roots (real and)
irrational
or for
4 ± 12
2 ± 3 or
obtained isw
2
M1
A2
or inspection eg (x − 1)(x2…..+ 1)
or equating coeffts with two correct
coeffts found
or (x − 2)2 = 3; may be implied by
correct roots or 12 obtained
[A1 for 12 and A1 for comment]
NB A2 is available only for correct
quadratic factor used; if wrong factor
used, allow A1 ft for obtaining two
irrational roots or for their
discriminant and comment re
irrational [no ft if their discriminant
is negative]
5
5
11
4751
Mark Scheme
June 2009
4751 (C1) Introduction to Advanced Mathematics
Section A
(0, 14) and (14/4, 0) o.e. isw
1
2
2 ( s − ut )
o.e. as final answer
t2
( s − ut ) ]
[condone [ a =]
0.5t 2
[ a =]
4
M2 for evidence of correct use of
gradient with (2, 6) eg sketch with
‘stepping’ or y − 6 = −4(x − 2) seen or y
= −4x + 14 o.e. or
M1 for y = −4x + c [accept any letter or
number] and M1 for 6 = −4 × 2 + c;
A1 for (0, 14) [c = 14 is not sufficient for
A1] and A1 for (14/4, 0) o.e.; allow
when x = 0, y = 14 etc isw
3
M1 for each of 3 complete correct
steps, ft from previous error if
equivalent difficulty [eg dividing by t
does not count as step – needs to be
by t2]
[ a =]
3
10 www
4
( s − ut )
1 2
2
t
gets M2 only (similarly
3
other triple-deckers)
M1 for f(3) = 1 soi and A1 for
31 − 3k = 1 or 27 − 3k = −3 o.e. [a
correct 3-term or 2-term equation]
3
long division used:
M1 for reaching (9 − k)x + 4 in working
and A1 for 4 + 3(9 − k) = 1 o.e.
4
5
x < 0 or x > 6 (both required)
(i) 10 www
2
2
equating coeffts method:
M2 for (x − 3)(x2 + 3x −1) [+ 1] o.e.
(from inspection or division)
3
B1 each;
if B0 then M1 for 0 and 6 identified;
2
M1 for
5× 4× 3
5× 4
or
or for
3 × 2 (×1)
2 (×1)
1 5 10
(ii) 80 www or ft 8 × their (i)
2
10
5
1 seen
B2 for 80x3; M1 for 23 or (2x)3 seen
4
16
1
4752
6
Mark Scheme
June 2009
any general attempt at n being odd
and n being even even
M1
M0 for just trying numbers, even if some
odd, some even
n odd implies n3 odd and odd − odd
= even
A1
n even implies n3 even and even −
even = even
A1
or n(n2 − 1) used with n odd implies n2 −
1 even and odd × even = even etc
[allow even × odd = even]
or A2 for n(n − 1)(n + 1) = product of 3
consecutive integers; at least one even
so product even;
odd3 − odd = odd etc is not sufft for A1
SC1 for complete general method for
only one of odd or even eg n = 2m
leading to 2(4m3 − m)
7
(i) 1
2
8
(ii) 1000
(i) 2/3 www
1
2
B1 for 50 or for 25 × 1/25 o.e.
3
M1 for 4/6 or for
48 = 2 12 or 4 3 or
27 = 3 3 or 108 = 3 12 or for
9
10
(ii) 43 − 30 2 www as final
answer
3
(i) (x + 3)2 − 4
3
(ii) ft their (−a, b);
if error in (i), accept (−3, −4) if
evidence of being independently
obtained
2
3
4
9
M2 for 3 terms correct of 25 − 15 2 −
15 2 + 18 soi, M1 for 2 terms correct
B1 for a = 3, B2 for b = − 4 or M1 for 5 −
32 soi
B1 each coord.; allow x = −3, y = −4; or
5
⎡ −3⎤
⎥ o.e. oe for sketch with −3
⎣ −4 ⎦
M1 for ⎢
(x2 − 9)(x2 + 4)
M2
x2 = 9 [or −4] or ft for integers
/fractions if first M1 earned
x = ±3 cao
M1
and −4 marked on axes but no coords
given
or correct use of quad formula or comp
sq reaching 9 and −4; allow M1 for
attempt with correct eqn at factorising
with factors giving two terms correct, or
sign error, or attempt at formula or
comp sq [no more than two errors in
formula/substn]; for this first M2 or M1
allow use of y etc or of x instead of x2
must have x2; or M1 for (x + 3)(x − 3);
this M1 may be implied by x = ±3
A0 if extra roots
if M0 then allow SC1 for use of factor
theorem to obtain both 3 and −3 as
roots or (x + 3) and (x − 3) found as
factors and SC2 for x2 + 4 found as
other factor using factor theorem [ie
max SC3]
A1
5
4
20
2
4752
Mark Scheme
Section B
11 i
y = 3x
ii
M1 for grad AB =
1− 3
or − 1/3 o.e.
6
eqn AB is y = -1/3 x + 3 o.e. or ft
M1
3x = -1/3x + 3 or ft
x = 9/10 or 0.9 o.e. cao
M1
A1
need not be simplified; no ft from
midpt used in (i); may be seen in (i)
but do not give mark unless used in
(ii)
eliminating x or y , ft their eqns
if find y first, cao for y then ft for x
A1
ft dep on both Ms earned
2
or square root of this; M1 for
y = 27/10 oe ft their 3 × their x
iii
2
June 2009
2
⎛9⎞
2
⎜ ⎟ (1 + 3 ) o.e and
⎝ 10 ⎠
2
2
4
2
⎛ 9 ⎞ ⎛ 27 ⎞
⎜ ⎟ + ⎜ ⎟ or 0.81 + 7.29 soi or ft
⎝ 10 ⎠ ⎝ 10 ⎠
completion to given answer
their coords (inc midpt)
or M1 for distance = 3 cos θ and tan
θ = 3 and M1 for showing
sin θ =
iv
v
2 10
2
9 www or ft their a 10
2
3
and completion
10
M1 for 62 + 22 or 40 or square roots
of these
2
2
M1 for ½ × 3 × 6 or
1
9
× their2 10 ×
10
2
10
2
12
3
4752
12
Mark Scheme
iA
expansion of one pair of brackets
M1
correct 6 term expansion
M1
June 2009
eg [(x + 1)](x2 − 6x + 8); need not be
simplified
eg x3 − 6x2 + 8x + x2 − 6x + 8;
or M2 for correct 8 term expansion:
x3 − 4x2 + x2 − 2x2 + 8x − 4x − 2x +
8, M1 if one error
allow equivalent marks working
backwards to factorisation, by long
division or factor theorem etc
or M1 for all three roots checked by
factor theorem and M1 for
comparing coeffts of x3
iB
cubic the correct way up
x-axis: −1, 2, 4 shown
y-axis 8 shown
G1
G1
G1
with two tps and extending beyond
the axes at ‘ends’
ignore a second graph which is a
translation of the correct graph
iC
ii
[y=](x − 2)(x − 5)(x − 7) isw or
(x − 3)3 − 5(x − 3)2 + 2(x − 3) + 8
isw or x3 − 14x2 + 59x − 70
2
M1 if one slip or for [y =] f(x − 3) or
for roots identified at 2, 5, 7
or for translation 3 to the left allow
M1 for complete attempt: (x + 4)( x +
1)(x − 1) isw or
(x + 3)3 − 5(x + 3)2 + 2(x + 3) + 8 isw
(0, −70) or y = −70
1
allow 1 for (0, −4) or y = −4 after f(x
+ 3) used
27 − 45 + 6 + 8 = −4 or 27 − 45 +
6 + 12 = 0
B1
or correct long division of x3 − 5x2 +
2x + 12 by (x − 3) with no remainder
or of x3 − 5x2 + 2x + 8 with rem −4
long division of f(x) or their f(x) + 4
by (x − 3) attempted as far as x3 −
3x2 in working
M1
or inspection with two terms correct
eg (x − 3)(x2 ………..− 4)
x2 − 2x − 4 obtained
A1
[ x =]
2±
2
( −2 )
2
− 4 × ( −4 )
2
or
M1
2
3
3
dep on previous M1 earned; for
attempt at formula or comp square
on their other ‘factor’
(x − 1) = 5
2 ± 20
o.e. isw or 1 ± 5
2
A1
5
13
4
4752
13
Mark Scheme
i
(5, 2)
20 or 2 5
ii
iii
iv
no, since 20 < 5 or showing
roots of y2 − 4y + 9 = 0 o.e. are
not real
y = 2x − 8 or simplified alternative
(x − 5)2 + (2x)2 = 20 o.e.
1
1
2
2
M1
5x2 − 10x + 5[ = 0] or better equiv. M1
obtaining x = 1 (with no other
roots) or showing roots equal
M1
one intersection [so tangent]
A1
(1, 4) cao
A1
alt method
y − 2 = −½ (x − 5) o.e.
2x + 2 − 2 = −½ (x − 5) o.e.
x=1
y = 4 cao
showing (1, 4) is on circle
M1
M1
A1
A1
B1
alt method
perp dist between y = 2x − 8 and
y = 2x + 2 = 10 cos θ where tan θ
=2
showing this is 20 so tgt
M1
x = 5 − 20 sin θ
x=1
(1, 4) cao
June 2009
0 for ± 20 etc
2
or ft from their centre and radius
M1 for attempt (no and mentioning
20 or 5) or sketch or solving by
formula or comp sq (−5)2 + (y − 2)2 =
20 [condone one error]
or SC1 for fully comparing distance
from x axis with radius and saying
yes
M1 for y − 2 = 2(x − 5) or ft from (i)
or M1 for y = 2x + c and subst their
(i)
or M1 for ans y = 2x + k, k ≠ 0 or −8
subst 2x + 2 for y [oe for x]
2
2
expanding brackets and rearranging
to 0; condone one error; dep on first
M1
o.e.; must be explicit; or showing line
joining (1,4) to centre is perp to y =
2x +2
allow y = 4
line through centre perp to y = 2x + 2
dep; subst to find intn with y = 2x + 2
by subst in circle eqn or finding dist
from centre = 20
[a similar method earns first M1 for
eqn of diameter, 2nd M1 for intn of
diameter and circle A1 each for x
and y coords and last B1 for showing
(1, 4) on line – award onlyA1 if (1, 4)
and (9, 0) found without (1, 4) being
identified as the soln]
M1
M1
A1
A1
or other valid method for obtaining x
allow y = 4
5
11
5
4751
Mark Scheme
4751 (C1) Introduction to Advanced Mathematics
1
3
[a =]2c 2 − b www o.e.
2
M1 for each of 3 complete correct
steps, ft from previous error if
equivalent difficulty
5 x − 3 < 2 x + 10
3 x < 13
13
x<
o.e.
3
M1
condone ‘=’ used for first two Ms
M0 for just 5x – 3 < 2(x + 5)
M1
or −13 < −3x or ft
M1
or ft; isw further simplification of 13/3;
M0 for just x < 4.3
1
allow y = 0, x = 4
bod B1 for x = 4 but do not isw:
0 for (0, 4) seen
0 for (4, 0) and (0, 10) both given
(choice) unless (4, 0) clearly identified
as the x-axis intercept
3
(i)
(4, 0)
3
(ii)
5x + 2(5 − x) = 20 o.e.
M1
for subst or for multn to make coeffts
same and appropriate addn/subtn;
condone one error
(10/3, 5/3) www isw
A2
or A1 for x = 10/3 and A1 for y = 5/3
o.e. isw; condone 3.33 or better and 1.67
or better
A1 for (3.3, 1.7)
4
(i)
translation
B1
0 for shift/move
by 
B1
or 4 units in negative x direction o.e.
sketch of parabola right way up and
with minimum on negative y-axis
B1
mark intent for both marks
min at (0, −4) and graph through −2
and 2 on x-axis
B1
must be labelled or shown nearby
 −4 
 or 4 [units] to left
 0
4
5
5
(ii)
(i)
(ii)
2
1
1
or ±
12
12
denominator = 18
(
numerator = 5 − 7 + 4 5 + 7
)
= 25 + 3 7 as final answer
M1 for
1
144 2
o.e. or for
144 = 12 soi
B1
B0 if 36 after addition
M1
for M1, allow in separate fractions
A1
allow B3 for
www
1
1
25 + 3 7
as final answer
18
4751
6
6
7
(i)
(ii)
Mark Scheme
cubic correct way up and with two
turning pts
B1
touching x-axis at −1, and through it at
2.5 and no other intersections
B1
y- axis intersection at −5
B1
2x3 − x2 − 8x − 5
January 2010
intns must be shown labelled or worked
out nearby
2
B1 for 3 terms correct or M1 for correct
expansion of product of two of the given
factors
attempt at f(−3)
−27 + 18 − 15 + k = 6
M1
A1
or M1 for long division by (x + 3) as far
as obtaining x2 −x and A1 for obtaining
remainder as k – 24 (but see below)
k = 30
A1
equating coefficients method:
M2 for (x + 3)(x2 – x + 8) [+6] o.e.
(from inspection or division) eg M2 for
obtaining x2 – x + 8 as quotient in
division
8
x3 + 15 x +
4
75 125
+ 3 www isw
x
x
B1 for both of x3 and
or x3 + 15x + 75x-1 + 125x-3 www isw
125
or 125x-3 isw
x3
and
M1 for 1 3 3 1 soi; A1 for each of 15x
and
75
or 75x-1 isw
x
or
SC2 for completely correct unsimplified
answer
2
4751
9
Mark Scheme
January 2010
x2 − 5x + 7 = 3x − 10
M1
or attempt to subst (y + 10)/3 for x
x2 − 8x + 17 [= 0] o.e or
y2− 4y + 13 [= 0] o.e
M1
condone one error; allow M1 for
x2 − 8x = −17 [oe for y] only if they go
on to completing square method
use of b2 − 4ac with numbers subst
(condone one error in substitution)
(may be in quadratic formula)
M1
or (x − 4)2 = 16 − 17 or (x − 4)2 + 1 = 0
(condone one error)
b2 − 4ac = 64 − 68 or −4 cao
[or 16 – 52 or −36 if y used]
A1
or (x − 4)2 = −1 or x = 4 ± −1
[< 0] so no [real] roots [so line and
curve do not intersect]
A1
[or (y – 2)2 = −9 or y = 2 ± −9 ]
or conclusion from comp. square; needs
to be explicit correct conclusion and
correct ft; allow ‘< 0 so no intersection’
o.e.; allow ‘−4 so no roots’ etc
allow A2 for full argument from sum of
two squares = 0; A1 for weaker correct
conclusion
some may use the condition b2 < 4ac for
no real roots; allow equivalent marks,
with first A1 for 64 < 68 o.e.
10 (i)
M1
grad CD =
5−3  2

= o.e. isw

3 − ( −1)  4

grad AB =
3 − ( −1)
4
or
isw
8
6 − ( −2 )
M1
same gradient so parallel www
A1
NB needs to be obtained independently
of grad AB
must be explicit conclusion mentioning
‘same gradient’ or ‘parallel’
if M0, allow B1 for 'parallel lines have
same gradient' o.e.
10 (ii)
[BC2=] 32 + 22
[BC2 =] 13
showing AD2 = 12 + 42 [=17] [≠BC2]
isw
M1
A1
A1
accept (6 − 3)2 + (3 − 5)2 o.e.
or [BC =] 13
or [AD =]
17
or equivalent marks for finding AD or
AD2 first
alt method: showing AC ≠ BD – mark
equivalently
3
4751
10 (iii)
10 (iv)
Mark Scheme
January 2010
[BD eqn is] y = 3
M1
eg allow for ‘at M, y = 3’ or for 3 subst
in eqn of AC
eqn of AC is y – 5 =6/5× (x – 3) o.e
[ y = 1.2x + 1.4 o.e.]
M2
or M1 for grad AC = 6/5 o.e. (accept
unsimplified) and M1 for using their
grad of AC with coords of A(−2, −1) or
C (3, 5) in eqn of line or M1 for
‘stepping’ method to reach M
M is (4/3, 3) o.e. isw
A1
allow : at M, x = 16/12 o.e. [eg =4/3] isw
A0 for 1.3 without a fraction answer
seen
midpt of BD = (5/2, 3) or equivalent
simplified form cao
M1
or showing BM ≠ MD oe
[BM = 14/3, MD = 7/3]
M1
or showing AM ≠ MC or AM2 ≠ MC2
A1
in these methods A1 is dependent on
coords of M having been obtained in
part (iii) or in this part; the coordinates
of M need not be correct; it is also
dependent on midpts of both AC and BD
attempted, at least one correct
midpt AC = (1/2, 2) or equivalent
simplified form cao
or ‘M is 2/3 of way from A to C’
conclusion ‘neither diagonal bisects
the other’
alt method: show that mid point of BD
does not lie on AC (M1) and vice-versa
(M1), A1 for both and conclusion
4
4751
Mark Scheme
11 (i)
centre C' = (3, −2)
radius 5
11 (ii)
showing (6 − 3)2 + (−6 + 2)2 = 25

1
1

 −3 
 o.e.
4
showing that AC ′ = C ′B = 
January 2010
0 for ±5 or −5
B1
interim step needed
B2
or B1 each for two of: showing
midpoint of AB = (3, −2); showing
B (0, 2) is on circle; showing AB = 10
or B2 for showing midpoint of
AB = (3, −2) and saying this is centre of
circle
or B1 for finding eqn of AB as
y = −4/3 x + 2 o.e. and B1 for finding
one of its intersections with the circle is
(0, 2)
or B1 for showing C′B = 5 and B1 for
showing AB = 10 or that AC′ and BC′
have the same gradient
or B1 for showing that AC′ and BC′
have the same gradient and B1 for
showing that B (0, 2) is on the circle
11 (iii)
grad AC' or AB = −4/3 o.e.
M1
or ft from their C', must be evaluated
grad tgt = −1/their AC' grad
M1
may be seen in eqn for tgt; allow M2 for
grad tgt = ¾ oe soi as first step
y −(−6) = their m(x − 6) o.e.
M1
or M1 for y = their m × x + c then subst
(6, −6)
y = 0.75x − 10.5 o.e. isw
A1
eg A1 for 4y = 3x – 42
allow B4 for correct equation www isw
11 (iv)
centre C is at (12, −14) cao
B2
B1 for each coord
circle is (x − 12)2 + (y + 14)2 = 100
B1
ft their C if at least one coord correct
5
4751
Mark Scheme
12 (i)
10
1
12 (ii)
[x =] 5 or ft their (i) ÷ 2
1
ht = 5[m] cao
1
12 (iii)
12 (iv)
January 2010
not necessarily ft from (i) eg they may
start again with calculus to get x = 5
d = 7/2 o.e.
M1
[y =] 1/5 × 3.5 × (10 − 3.5) o.e. or ft
M1
or ft their (ii) − 1.5 or their (i) ÷ 2 − 1.5
o.e.
or 7 – 1/5 × 3.52 or ft
= 91/20 o.e. cao isw
A1
or showing y – 4 = 11/20 o.e. cao
4.5 = 1/5 × x(10 − x) o.e.
M1
22.5 = x(10 − x) o.e.
M1
eg 4.5 = x(2 – 0.2x) etc
2x2 − 20x + 45 [= 0] o.e. eg
x2 – 10x + 22.5 [=0] or (x – 5)2 = 2.5
A1
cao; accept versions with fractional
coefficients of x2, isw
M1
or x − 5 = [ ± ] 2.5 o.e.; ft their
[ x =]
20 ± 40
1
or 5 ±
10 o.e.
4
2
width =
quadratic eqn provided at least M1
gained already; condone one error in
formula or substitution; need not be
simplified or be real
10 o.e. eg 2 2.5 cao
A1
6
accept simple equivalents only
GCE
Mathematics (MEI)
Advanced Subsidiary GCE 4751
Introduction to Advanced Mathematics (C1)
Mark Scheme for June 2010
Oxford Cambridge and RSA Examinations
4751
Mark Scheme
SECTION A
1
y = 3x + c or y − y1 = 3(x − x1)
2
3
June 2010
M1
allow M1 for 3 clearly stated/ used as
gradient of required line
y − 5 = their m(x − 4) o.e.
M1
or (4, 5) subst in their y = mx + c;
allow M1 for y − 5 = m(x − 4) o.e.
y = 3x − 7 or simplified equiv.
A1
condone y = 3x + c and c = −7
or B3 www
(i) 250a6b7
2
(ii) 16 cao
1
(iii) 64
2
ac =
ac + 5 =
y −5
o.e.
M1
y
o.e.
M1
[ y =] ( ac + 5 ) o.e. isw
B1 for two elements correct; condone
multiplication signs left in
SC1 for eg 250 + a6 + b7
condone ±64
M1 for [±]43 or for
only −64
4096 or for
M1 for each of 3 correct or ft correct
steps s.o.i. leading to y as subject
2
M1
or some/all steps may be combined;
2
allow B3 for [ y =] ( ac + 5 ) o.e. isw
or B2 if one error
4
(i)
2 − 2x > 6x + 5
M1
or 1 − x > 3x + 2.5
−3 > 8x o.e. or ft
M1
x < −3/8 o.e. or ft isw
M1
for collecting terms of their
inequality correctly on opposite sides
eg −8x > 3
allow B3 for correct inequality found
after working with equation
allow SC2 for −3/8 o.e. found with
equation or wrong inequality
4
(ii)
−4 < x < ½ o.e.
2
accept as two inequalities
M1 for one ‘end’ correct or for −4
and ½
5
(i)
7 3
2
M1 for
3
48 = 4 3 or
27 = 3 3
4751
5
Mark Scheme
(ii)
10 + 15 2
www isw
7
3
June 2010
B1 for 7 [B0 for 7 wrongly obtained]
and B2 for 10 + 15 2 or B1 for one
term of numerator correct;
if B0, then M1 for attempt to
multiply num and denom by 3 + 2
6
5 + 2k soi
M1
k = 12
A1
attempt at f(3)
M1
27 + 36 + m = 59 o.e.
A1
m = −4 cao
A1
allow M1 for expansion with 5x3 +
2kx3 and no other x3 terms
or M1 for (29 − 5) / 2 soi
must substitute 3 for x in cubic not
product
or long division as far as obtaining x2
+ 3x in quotient
or from division m − (−63) = 59 o.e.
or for 27 + 3k + m = 59 or ft their k
7
1 + 2x + 32 x 2 + 12 x 3 + 161 x 4 oe (must
be simplified) isw
4
B3 for 4 terms correct, or B2 for 3
terms correct or for all correct but
unsimplified (may be at an earlier
stage, but factorial or nCr notation
must be expanded/worked out)
or B1 for 1, 4, 6, 4, 1 soi or for
1 + ... + 161 x 4 [must have at least one
other term]
8
5(x + 2)2 − 14
4
B1 for a = 5, and B1 for b = 2
and B2 for c = −14 or M1 for c = 6 −
their ab2 or
M1 for [their a](6/their a − their b2)
[no ft for a = 1]
9
mention of −5 as a square root of
25 or (−5)2 = 25
M1
condone −52 = 25
−5 −5 ≠ 0 o.e.
or x + 5 = 0
M1
or, dep on first M1 being obtained,
allow M1 for showing that 5 is the
only soln of x − 5 = 0
allow M2 for
x2 − 25 = 0
(x + 5)(x − 5) [= 0]
so x − 5 = 0 or x + 5 = 0
Section A Total: 36
4751
Mark Scheme
June 2010
SECTION B
10
10
(i)
(ii)
(2x − 3)(x + 1)
M2
M1 for factors with one sign error
or giving two terms correct
allow M1 for 2(x − 1.5)(x + 1) with
no better factors seen
x = 3/2 and − 1 obtained
B1
or ft their factors
graph of quadratic the correct way
up and crossing both axes
B1
crossing x-axis only at 3/2 and − 1 B1
or ft from their roots in (i), or their
factors if roots not given
crossing y-axis at −3
10
10
(iii) use of b2 − 4ac with numbers
subst (condone one error in
substitution) (may be in quadratic
formula)
(iv)
for x = 3/2 condone 1 and 2 marked
on axis and crossing roughly
halfway between;
intns must be shown labelled or
worked out nearby
B1
M1
may be in formula
or (x – 2.5)2 =6.25 − 10 or (x – 2.5)2 +
3.75 = 0 oe (condone one error)
25 − 40 < 0 or −15 obtained
A1
or −15 seen in formula
or (x – 2.5)2 = −3.75 oe
or x = 2.5 ± −3.75 oe
2x2 − x − 3 = x2 − 5x + 10 o.e.
M1
attempt at eliminating y by subst or
subtraction
x2 + 4x − 13 [= 0]
M1
or (x + 2)2 = 17; for rearranging to
form ax 2 + bx + c [= 0] or to
completing square form
condone one error for each of 2nd
and 3rd M1s
use of quad. formula on resulting
eqn (do not allow for original
quadratics used)
M1
or x + 2 = ± 17 o.e.
2nd and 3rd M1s may be earned for
good attempt at completing square
as far as roots obtained
−2 ± 17 cao
A1
4751
11
11
Mark Scheme
(i)
(ii)
June 2010
1− 3
[= −1/3]
5 − (−1)
y − 3 = their grad (x – (−1)) or
y − 1 = their grad (x − 5)
M1
y = −1/3x + 8/3 or 3y = −x + 8 o.e
isw
A1
when y = 0, x = 8; when x = 0,
y = 8/3 or ft their (i)
M1
allow y = 8/3 used without
explanation if already seen in eqn in
(i)
[Area =] ½ × 8/3 × 8 o.e. cao isw
M1
NB answer 32/3 given;
allow 4 × 8/3 if first M1 earned;
or
M1 for
grad AB =
M1
or use of y = their gradient x + c
with coords of A or B
y − 3 x − ( −1)
=
o.e.
or M2 for
1 − 3 5 − ( −1)
o.e. eg x + 3y − 8 = 0 or 6y = 16 −
2x
allow B3 for correct eqn www
2
0  13 (8 − x )dx =  13 (8x − 12 x ) 0
8
8
and M1 dep for
11
(iii) grad perp = −1/grad AB stated, or
used after their grad AB stated in
this part
midpoint [of AB] = (2, 2)
1
3
( 64 − 32[−0])
M1
or showing 3 × −1/3 = −1
if (i) is wrong, allow the first M1
here ft, provided the answer is
correct ft
M1
must state ‘midpoint’ or show
working
y − 2 = their grad perp (x − 2) or ft M1
their midpoint
for M3 this must be correct, starting
from grad AB = −1/3, and also
needs correct completion to given
ans y = 3x − 4
alt method working back from
ans:
or
mark one method or the other, to
benefit of candidate, not a mixture
grad perp = −1/grad AB and
showing/stating same as given
line
M1
eg stating − 1/3 × 3 = −1
finding intn of their
y = −1/3x − 8/3and y = 3x − 4 is
(2, 2)
M1
or showing that (2, 2) is on y = 3x −
4, having found (2, 2) first
showing midpt of AB is (2, 2)
M1
[for both methods: for M3 must be
fully correct]
4751
11
12
Mark Scheme
(iv)
(i)
June 2010
subst x = 3 into y = 3x − 4 and
obtaining centre = (3, 5)
M1
or using (−1 − 3)2 + (3 − b)2 = (5 −
3)2 + (1 − b)2 and finding (3, 5)
r2 = (5 − 3)2 + (1 − 5)2 o.e.
M1
or (−1 − 3)2 + (3 − 5)2 or ft their
centre using A or B
r = 20 o.e. cao
A1
eqn is (x − 3)2 + (y − 5)2 = 20 or ft
their r and y-coord of centre
B1
condone (x − 3)2 + (y − b)2 = r2 o.e.
or (x − 3)2 + (y – their 5)2 = r2 o.e.
(may be seen earlier)
trials of at calculating f(x) for at
least one factor of 30
M1
M0 for division or inspection used
details of calculation for f(2) or
f(−3) or f(−5)
A1
attempt at division by (x − 2) as
far as x3 − 2x2 in working
M1
correctly obtaining x2 + 8x + 15
A1
factorising a correct quadratic
factor
M1
for factors giving two terms of
quadratic correct; M0 for formula
without factors found
(x − 2)(x + 3)(x + 5)
A1
condone omission of first factor
found; ignore ‘= 0’ seen
or equiv for (x + 3) or (x + 5); or
inspection with at least two terms of
quadratic factor correct
or B2 for another factor found by
factor theorem
allow last four marks for
(x − 2)(x + 3)(x + 5) obtained; for
all 6 marks must see factor theorem
use first
12
(ii)
sketch of cubic right way up, with
two turning points
B1
0 if stops at x-axis
values of intns on x axis shown,
correct (−5, −3, and 2) or ft from
their factors/ roots in (i)
B1
on graph or nearby in this part
y-axis intersection at −30
B1
mark intent for intersections with
both axes
or x = 0, y = −30 seen in this part if
consistent with graph drawn
4751
12
Mark Scheme
(iii) (x − 1) substituted for x in either
form of eqn for y = f(x)
June 2010
M1
correct or ft their (i) or (ii) for
factorised form; condone one error;
allow for new roots stated as −4,−2
and 3 or ft
(x − 1)3 expanded correctly (need
not be simplified) or two of their
factors multiplied correctly
M1
dep
or M1 for correct or correct ft
multiplying out of all 3 brackets at
once, condoning one error [x3 − 3x2
+ 4x2 + 2x2 + 8x − 6x − 12x − 24]
correct completion to given
answer
[condone omission of ‘y =’]
M1
unless all 3 brackets already
expanded, must show at least one
further interim step
allow SC1 for (x + 1) subst and
correct exp of (x + 1)3 or two of
their factors ft
or, for those using given answer:
M1 for roots stated or used as
−4,−2 and 3 or ft
A1 for showing all 3 roots satisfy
given eqn
B1 for comment re coefft of x3 or
product of roots to show that eqn of
translated graph is not a multiple of
RHS of given eqn
Section B Total: 36
GCE
Mathematics (MEI)
Advanced Subsidiary GCE
Unit 4751: Introduction to Advanced Mathematics
Mark Scheme for January 2011
Oxford Cambridge and RSA Examinations
4751
Mark Scheme
January 2011
Marking instructions for GCE Mathematics (MEI): Pure strand
1.
You are advised to work through the paper yourself first. Ensure you familiarise yourself
with the mark scheme before you tackle the practice scripts.
2.
You will be required to mark ten practice scripts. This will help you to understand the mark
scheme and will not be used to assess the quality of your marking. Mark the scripts
yourself first, using the annotations. Turn on the comments box and make sure you
understand the comments. You must also look at the definitive marks to check your
marking. If you are unsure why the marks for the practice scripts have been awarded in
the way they have, please contact your Team Leader.
3.
When you are confident with the mark scheme, mark the ten standardisation scripts. Your
Team Leader will give you feedback on these scripts and approve you for marking. (If your
marking is not of an acceptable standard your Team Leader will give you advice and you
will be required to do further work. You will only be approved for marking if your Team
Leader is confident that you will be able to mark candidate scripts to an acceptable
standard.)
4.
Mark strictly to the mark scheme. If in doubt, consult your Team Leader using the
messaging system within scoris, by email or by telephone. Your Team Leader will be
monitoring your marking and giving you feedback throughout the marking period.
An element of professional judgement is required in the marking of any written paper.
Remember that the mark scheme is designed to assist in marking incorrect solutions.
Correct solutions leading to correct answers are awarded full marks but work must not be
judged on the answer alone, and answers that are given in the question, especially, must
be validly obtained; key steps in the working must always be looked at and anything
unfamiliar must be investigated thoroughly.
Correct but unfamiliar or unexpected methods are often signalled by a correct result
following an apparently incorrect method. Such work must be carefully assessed. When a
candidate adopts a method which does not correspond to the mark scheme, award marks
according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if
several marks or candidates are involved) you should contact your Team Leader.
5.
The following types of marks are available.
M
A suitable method has been selected and applied in a manner which shows that the
method is essentially understood. Method marks are not usually lost for numerical errors,
algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to
indicate an intention of using some method or just to quote a formula; the formula or idea
must be applied to the specific problem in hand, eg by substituting the relevant quantities
into the formula. In some cases the nature of the errors allowed for the award of an M mark
may be specified.
A
Accuracy mark, awarded for a correct answer or intermediate step correctly obtained.
Accuracy marks cannot be given unless the associated Method mark is earned (or
implied). Therefore M0 A1 cannot ever be awarded.
B
Mark for a correct result or statement independent of Method marks.
1
4751
Mark Scheme
January 2011
E
A given result is to be established or a result has to be explained. This usually requires
more working or explanation than the establishment of an unknown result.
Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong
working following a correct form of answer is ignored. Sometimes this is reinforced in the
mark scheme by the abbreviation isw. However, this would not apply to a case where a
candidate passes through the correct answer as part of a wrong argument.
6.
When a part of a question has two or more ‘method’ steps, the M marks are in principle
independent unless the scheme specifically says otherwise; and similarly where there are
several B marks allocated. (The notation ‘dep *’ is used to indicate that a particular mark is
dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may
happen that when a candidate has once gone wrong in a part of a question, the work from
there on is worthless so that no more marks can sensibly be given. On the other hand,
when two or more steps are successfully run together by the candidate, the earlier marks
are implied and full credit must be given.
7.
The abbreviation ft implies that the A or B mark indicated is allowed for work correctly
following on from previously incorrect results. Otherwise, A and B marks are given for
correct work only – differences in notation are of course permitted. A (accuracy) marks are
not given for answers obtained from incorrect working. When A or B marks are awarded
for work at an intermediate stage of a solution, there may be various alternatives that are
equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark
scheme rationale. If this is not the case please consult your Team Leader.
Sometimes the answer to one part of a question is used in a later part of the same
question. In this case, A marks will often be ‘follow through’. In such cases you must
ensure that you refer back to the answer of the previous part question even if this is not
shown within the image zone. You may find it easier to mark follow through questions
candidate-by-candidate rather than question-by-question.
8.
Wrong or missing units in an answer should not lead to the loss of a mark unless the
scheme specifically indicates otherwise. Candidates are expected to give numerical
answers to an appropriate degree of accuracy, with 3 significant figures often being the
norm. Small variations in the degree of accuracy to which an answer is given (eg 2 or 4
significant figures where 3 is expected) should not normally be penalised, while answers
which are grossly over- or under-specified should normally result in the loss of a mark.
The situation regarding any particular cases where the accuracy of the answer may be a
marking issue should be detailed in the mark scheme rationale. If in doubt, contact your
Team Leader.
9.
Rules for crossed out and/or replaced work
If work is crossed out and not replaced, examiners should mark the crossed out work if it is
legible.
If a candidate attempts a question more than once, and indicates which attempt he/she
wishes to be marked, then examiners should do as the candidate requests.
If two or more attempts are made at a question, and just one is not crossed out, examiners
should ignore the crossed out work and mark the work that is not crossed out.
If there are two or more attempts at a question which have not been crossed out,
examiners should mark what appears to be the last (complete) attempt and ignore the
others.
2
4751
Mark Scheme
January 2011
NB Follow these maths-specific instructions rather than those in the assessor handbook.
10.
For a genuine misreading (of numbers or symbols) which is such that the object and the
difficulty of the question remain unaltered, mark according to the scheme but following
through from the candidate’s data. A penalty is then applied; 1 mark is generally
appropriate, though this may differ for some units. This is achieved by withholding one A
mark in the question.
Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.
11.
Annotations should be used whenever appropriate during your marking.
The A, M and B annotations must be used on your standardisation scripts for
responses that are not awarded either 0 or full marks. It is vital that you annotate
standardisation scripts fully to show how the marks have been awarded.
For subsequent marking you must make it clear how you have arrived at the mark you
have awarded.
12.
For answers scoring no marks, you must either award NR (no response) or 0, as follows:
Award NR (no response) if:

Nothing is written at all in the answer space

There is a comment which does not in any way relate to the question being asked
(“can’t do”, “don’t know”, etc.)

There is any sort of mark that is not an attempt at the question (a dash, a question
mark, etc.)
The hash key [#] on your keyboard will enter NR.
Award 0 if:

There is an attempt that earns no credit. This could, for example, include the
candidate copying all or some of the question, or any working that does not earn any
marks, whether crossed out or not.
13.
The following abbreviations may be used in this mark scheme.
M1
A1
B1
E1
U1
G1
M1 dep*
cao
ft
isw
oe
rot
sc
soi
www
method mark (M2, etc, is also used)
accuracy mark
independent mark
mark for explaining
mark for correct units
mark for a correct feature on a graph
method mark dependent on a previous mark, indicated by *
correct answer only
follow through
ignore subsequent working
or equivalent
rounded or truncated
special case
seen or implied
without wrong working
3
4751
14.
Mark Scheme
January 2011
Annotating scripts. The following annotations are available:
and 
BOD
Benefit of doubt
FT
Follow through
ISW
Ignore subsequent working (after correct answer obtained)
M0, M1
Method mark awarded 0, 1
A0, A1
Accuracy mark awarded 0, 1
B0, B1
Independent mark awarded 0,1
SC
Special case
^
Omission sign
MR
Misread
Highlighting is also available to highlight any particular points on a script.
15.
The comments box will be used by the Principal Examiner to explain his or her marking of
the practice scripts for your information. Please refer to these comments when checking
your practice scripts.
Please do not type in the comments box yourself. Any questions or comments you
have for your Team Leader should be communicated by the scoris messaging system, email or by telephone.
16.
Write a brief report on the performance of the candidates. Your Team Leader will tell you
when this is required. The Assistant Examiner’s Report Form (AERF) can be found on the
Cambridge Assessment Support Portal. This should contain notes on particular strengths
displayed, as well as common errors or weaknesses. Constructive criticisms of the
question paper/mark scheme are also appreciated.
17.
Link Additional Objects with work relating to a question to those questions (a chain link
appears by the relevant question number) – see scoris assessor Quick Reference Guide
page 19-20 for instructions as to how to do this – this guide is on the Cambridge
Assessment Support Portal and new users may like to download it with a shortcut on your
desktop so you can open it easily! For AOs containing just formulae or rough working not
attributed to a question, tick at the top to indicate seen but not linked. When you submit
the script, scoris asks you to confirm that you have looked at all the additional objects.
Please ensure that you have checked all Additional Objects thoroughly.
18.
The schedule of dates for the marking of this paper is displayed under ‘OCR Subject
Specific Details’ on the Cambridge Assessment Support Portal. It is vitally important that
you meet these requirements. If you experience problems that mean you may not be able
to meet the deadline then you must contact your Team Leader without delay.
4
4751
Mark Scheme
January 2011
SECTION A
1
y = 5x + 3
3
M2 for y − 13 = 5(x − 2) oe
or M1 for y = 5x [+ k] [k = letter or
number other than −4] and M1 for
13 = their m × 2 + k
isw attempted conversion of 1/16 to
decimals
or M1 for y  b = 5(x  a) with wrong a, b or for
y − 13 = their 5(x − 2) oe
M0 for first M if 1/5 used as gradient even if 5 seen
first; second M still available if earned
2
(i)(A) 1/16
1
accept 0.0625
2
(i)(B) 1
1
2
(ii) 256/625
2
M1 for num or denom correct or for 4/5
or 0.8
accept 0.4096
3
9 y10
oe as final answer
2 x2
3
1 for each ‘term’; 27/6 gets 0 for first
term
allow eg 4.5x-2y10
set image ‘fit to height’ so that in marking this question
you also check that there is no working on the back
page attached to the image
if 0, allow B1 for (3xy4)3 = 27x3y12
4
x > 5/2 oe (5/2 oe not sufft)
2
M1 for 5 < 2x or for 5/2 oe obtained
with equation or wrong inequality
5
M0 for just 2x <  5 (not sufft) ; M1 for x > 5/2
4751
5
Mark Scheme
3V
 l2  r2
 r2
January 2011
M1
for correctly getting non- ‘l2  r2’ terms
on other side[M0 for ‘triple decker’
fraction]
may be done in several steps, if so, condone omission
of brackets in eg 9V2 = π2r4 l2  r2 if they recover – if
not, do not give 1st M1 [but can earn the 2nd M1]
M1
oe or ft; for squaring correctly
for combined steps, allow credit for correct process
where possible;
M1
oe or ft; for getting l term as subject
eg π2r4l2 as the term on one side
M1
oe. or ft; mark final answer; for finding
square root ( and dealing correctly with
coefficient of l term if needed at this
stage); condone ±√etc
For M4, the final expression must be totally correct,
[condoning omission of l and insertion of ±]
B3 for all correct except for sign
error(s)
B2 for 2 terms correct numerically,
ignoring any sign error or for 32, 240
and 720 found
or B2 for all correct, including signs,
but unsimplified
B1 for binomial coeffts 1, 5, 10 used or
1 5 10 10 5 1 seen
accept terms listed separately; condone 240x1
2
 3V 
2
2
 2  l r

r


2
 3V 
l2   2   r2
r 
2
 3V 
[l ]  2   r 2
r 
6
32 − 240x + 720x2 isw
4
SC3 for − 240x + 720x2  1080x3 isw or
for 243x5+ 810x4 1080x3
or SC2 for these terms with sign
error(s)
6
eg M4 for
9V 2   2 r 6
 r2
expressions left in nC r form or with factorials not sufft
4751
7
Mark Scheme
(i) 37/2 oe or k = 7/2 oe
2
January 2011
M0 for just 81 = 3 × 3 × 3 × 3 oe – indices needed
34
81
or 1/2 or 81 31/2 or 33 3
3
3
allow an M mark for partially correct work still seen in
1/2
or 27 × 3 or better or for 81 = 34 or 3
34
1
fraction form eg 1/2 gets mark for 81 = 34
1/2
1/2
3
or (following
= 3 or
3
3
correct rationalisation of denominator)
for 27 = 33
M1 for
isw conversion of 7/2 oe
7
8
(ii)
3
14  5 3
28  10 3
or
www isw
11
22
(7/11, 24/11) oe www
3
M1 for multiplying num and denom by
5 + 3
and M1 for num or denom correct in
final answer (M0 if wrongly obtained)
2nd M1 is not dependent on 1st M1
B2 for one coord correct; condone not
expressed as coords, isw
or M1 for subst or elimination; eg x +
2(5x − 1) = 5 oe; condone one error
SC2 for mixed fractions and decimals
eg (3.5/5.5, 12/5.5)
9
(i) ½ × 2x × (x + 2 + 3x + 6) oe
M1
x(4x + 8) = 140 oe and given ans
x2 + 2x − 35 = 0 obtained correctly
with at least one further interim step
A1
correct statement of area of trap; may be eg 2x(x + 2) + ½ × 2x × (2x + 4)
rectangle ± triangle, or two triangles
condone missing brackets for M1; condone also for A1
if expansion is treated as if they were there
7
4751
Mark Scheme
(ii) [AB =] 21 www
3
or B2 for x = [−7 or] 5 cao www or for
AB = 21 or 15
or M1 for (x + 7)(x − 5) [= 0]or formula
or completing square used eg (x + 1)2 
36 [= 0]; condone one error eg factors
with sign wrong or which give two
terms correct when expanded
January 2011
may be done in (i) if not here – allow the marks if seen
in either part of the image – some candidates are
omitting the request in (i) and going straight to solving
the equation (in which case give 0 [not NR] for (i), but
annotate when the image appears again in (ii))
5 on its own or AB = 5 with no working scores 0; we
need to see x = 5
or M1 for showing f(5) = 0 without
stating x = 5
10
(i) P  Q
1
(ii) none [of the above]
1
(iii) P  Q
1
or  or ‘Q  P’
Condone single arrows
or 
Section A Total: 36
8
4751
Mark Scheme
SECTION B
11
06
oe [= −3] isw
(i) grad AB =
1   1
grad BC =
04
oe [= 1/3] isw
1  13
product of grads = −1 [so lines perp]
stated or shown numerically
M1
for full marks, it should be clear that
grads are independently obtained
eg grads of 3 and 1/3 without earlier working earn
M1M0
or ‘one grad is neg. reciprocal of other’
for M3, must be fully correct, with gradients evaluated
at least to 6/2 and 4/12 stage
M1
M1
or
M1 for length of one side (or square of
it)
M1 for length of other two sides (or
their squares) found independently
M1 for showing or stating that Pythag
holds [so triangle rt angled]
11
(ii) AB = 40 or BC = 160
January 2011
M1
AB2 = 62 + 22 = 40, BC2 = 42 + 122 = 160, AC2 = 142
+ 22 = 200
allow M1 for
(1   1) 2  (6  0) 2 or for
(13  1) 2  (4  0) 2
½ × 40 × 160 oe or ft their AB, BC
M1
40
A1
or M1 for one of area under AC (=70),
under AB (=6) under BC (=24) (accept
unsimplified) and M1 for their trap. −
two triangles
9
or for rectangle  3 triangles method,
1
1
1
[ 6  14  (2)(6)  (4)(12)  (2)(14)
2
2
2
=84  6  24  14]
M1 for two of the 4 areas correct and M1 for the
subtraction
4751
11
Mark Scheme
January 2011
condone ‘AB and BC are perpendicular’ or ‘ABC is
right angled triangle’ provided no spurious extra
reasoning
(iii) angle subtended by diameter =
90° soi
B1
or angle at centre = twice angle at
circumf = 2 × 90 = 180 soi
or showing BM = AM or CM, where M
is midpt of AC; or showing that BM =
½ AC
mid point M of AC = (6, 5)
B2
allow if seen in circle equation ; M1 for
correct working seen for both coords
M1
accept unsimplified; or eg r2 = 72 + 12
or 52 + 52; may be implied by correct
equation for circle or by correct method
for AM, BM or CM ft their M
allow M1 bod intent for AC =
100
or x 2  y 2  12 x  10 y  11  0
must be simplified (no surds)
No section to be ruled; no curving back; condone slight
‘flicking out’ at ends; condone some doubling (eg
erased curves may continue to show)
rad of circle =
1
2
142  22    12 200
2
oe or equiv using r
(x − a)2 + (y − b)2 = r2 seen or
(x – their 6)2 + (y – their 5)2 = k used,
with k > 0
200 followed by r =
M1
(x − 6)2 + (y − 5)2 = 50 cao
A1
11
(iv) (11, 10) cao
1
12
(i)(A) sketch of cubic correct way up
and with two tps, crossing x-axis in 3
distinct points
B1
0 if stops at x-axis; condone not
crossing y-axis
crossing x axis at 1, 2.5 and 4
B1
intersections labelled on graph or shown allow 2.5 indicated by graph crossing halfway between
their marked 2 and 3 on scale; allow if no graph but 0
nearby in this part; mark intent for
if graph inconsistent with values
intersections with both axes (eg
condone graphs stopping at axes)
crossing y axis at −20
B1
or x = 0, y = −20 seen in this part if
consistent with graph drawn
10
allow if no graph, but eg B0 for graph with intn on +ve
y-axis or nowhere near their indicated 20
4751
12
Mark Scheme
(i)(B) correct expansion of two
brackets
M1
or M2 for all 3 brackets multiplied at
once, showing all 8 terms (M1 if error
in one term): 2x3 − 8x2 − 2x2 − 5x2 + 8x
+ 5x + 20x – 20
January 2011
eg M1 for (2x  5)(x2  5x + 4)
condone missing brackets if intent clear /used correctly
correct interim step(s) multiplying out M1
linear and quadratic factors before
given answer
12
12
or
showing that 1, 2.5 and 4 all satisfy
f(x) = 0 for cubic in 2x3 ... form
or
comparing coeffts of eg x3 in the two
forms
M1
(ii)(A) 250 − 375 + 165 − 40 isw
B1
M1
or
M1 for dividing 2x3 ... form by one of
the linear factors and M1 for factorising
the resultant quadratic factor
‘2 × 125 + 15 × 25 + 33 × 5  40’ is not sufft
or
showing that x  5 is a factor by eg
division and then stating that x = 5 is
root or that g(5) = 0
or
[g(5) =] f(5)  20 = 5 × 4 × 1  20 [= 0]
may be in attempt at division
allow if seen in (ii)(A)
division by (x − 5) as far as 2x3 − 10x2 M1
seen in working
or inspection/equating coefficients with
two terms correct eg (2x2 ….. + 8)
for division: condone signs of 2x3 − 10x2 changed for
subtraction, or subtraction sign in front of first term
2x2 − 5x + 8 obtained isw
eg may be seen in grid;
(ii) (B) (x − 5) seen or used as linear
factor
M1
A1
condone g(x) not expressed as product
11
4751
12
Mark Scheme
(ii)(C) b2 − 4ac used on their
quadratic factor
M1
may be in formula
(5)2 − 4 × 2 × 8 oe and negative [or
−39] so no [real] root [may say only
one [real] root, thinking of x = 5]
A1
[or allow 2 marks for complete correct
attempt at completing square and
conclusion, or using calculus to show
min value is above x-axis and comment
re curve all above x-axis]
January 2011
no ft for A mark from wrong quadratic factor
condone error in working out 39 if correct
unsimplified expression seen and neg result obtained
52  4 × 2 × 8 evaluated correctly with comment is
eligible for A1, otherwise bod for the M1 only
12
13
(iii) translation
B1
NB ‘Moves’ not sufficient for this first
mark
 0 


 20 
B1
or 20 down;
(i) (0, −2) or ‘crosses y-axis at −2’ oe
isw
B1
1
4
( 2 , 0) oe isw
B2
B0 for second mark if choice of one wrong, one right
description
condone y = 2
or [when y = 0],
1
4
[ x ]  2 or 
2 or  4 2 isw
B1 for one root correct
12
4751
13
Mark Scheme
January 2011
(ii) [y = ] x2 = x4 − 2 oe and
rearrangement to
x4 − x2 − 2 [= 0] or y2 − y − 2 [=0]
M1
(x2 − 2)(x2 + 1) = 0 oe in y
M1
or formula or completing square;
if completing square, and haven’t arranged to zero, can
condone one error; condone
earn first M1 as well for an attempt such as
2
replacement of x by another letter or by (x2  0.5)2 = 2.25
x for 2nd M1 (but not the 3rd M1)
x2 = 2 [or −1] or y = 2 or −1 or ft
or x  2 or x   2 or ft
M1
dep on 2nd M1; allow inclusion of
correct complex roots; M0 if any
incorrect roots are included for x2 or x
NB for second and third M: M0 for x2  2 = 0 or x2 = 2
oe straight from quartic eqn – some candidates
probably thinking x4  x2 simplifies to x2; last two
marks for roots are available as B marks
( 2 , 2) and (− 2 , 2); with no other
intersections given
B2
or B1 for one of these two intersections
(even if extra intersections given) or for
x   2 (and no other roots) or for y =
2 (and no other roots), marking to
candidates’ advantage
some candidates having several attempts at solving this
equation – mark the best in this particular case
13
4751
13
Mark Scheme
(iii) from x4 − kx2 − 2 [= 0]:
Allow x2 replaced by other letters or x
or from y2 – k2y – 2k2 [= 0]
k2 + 8 > 0 oe
B1
k4 + 8k2 > 0 oe
k  k 2  8  0 for all k
B1
k 2  k 4  8k 2 > 0 oe for all k
[so there is a positive root for x2 and
hence real root for x and so
intersection]
[so there is a positive root for y and
hence real root for x and so intersection]
if B0B0, allow SC1 for
k  k2  8
or
2
k 2  k 4  8k 2
obtained [need not be
2
simplified]
Section B Total: 36
14
January 2011
[alt methods: may use completing square to show
similarly, or comment that at x = 0 the quadratic is
above the quartic and that as x → ∞, x4 – 2 > kx2 for all
k]
condone lack of brackets in (–k)2
GCE
Mathematics (MEI)
Advanced Subsidiary GCE
Unit 4751: Introduction to Advanced Mathematics
Mark Scheme for June 2011
Oxford Cambridge and RSA Examinations
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support which keep pace with the changing needs of today’s society.
This mark scheme is published as an aid to teachers and students, to indicate the requirements
of the examination. It shows the basis on which marks were awarded by Examiners. It does not
indicate the details of the discussions which took place at an Examiners’ meeting before marking
commenced.
All Examiners are instructed that alternative correct answers and unexpected approaches in
candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills
demonstrated.
Mark schemes should be read in conjunction with the published question papers and the Report
on the Examination.
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© OCR 2011
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4751
Mark Scheme
SECTION A
1
x > −13/4 o.e. isw www
3
condone x > 13/−4 or 13/−4 < x;
M2 for 4x > −13 or M1 for one side of
this correct with correct inequality,
and B1 for final step ft from their ax > b
or c > dx for a ≠ 1and d ≠ 1;
June 2011
M1 for 13 > −4x (may be followed by 13/−4 > x, which
earns no further credit);
6x + 3 > 2x + 5 is an error not an MR; can get M1 for
4x > ... following this, and then a possible B1
if no working shown, allow SC1 for
−13/4 oe with equals sign or wrong
inequality
2
7
2
condone y = 7 or (5, 7);
condone omission of brackets;
k − (−5)
= 3 or other correct
5 −1
use of gradient eg triangle with 4
across, 12 up
or M1 for correct method for eqn of line and
x = 5 subst in their eqn and evaluated to find k;
condone ±4/3;
M1 for just −4/3;
M1 for
3
(i) 4/3 isw
2
or M1 for both of y − k = 3(x − 5) oe and
y − (−5) = 3(x − 1) oe
M1 for numerator or denominator
correct or for
3
1
or
oe or for
4
3
 
4
1
2
 16 
  soi
9
1
allow M1 for 16 = 4 and 9 = 3 soi;
condone missing brackets
4751
3
Mark Scheme
(ii)
2a
or 2ac −5
c5
3
B1 for each ‘term’ correct;
mark final answer;
June 2011
1
condone a ;
condone multiplication signs but 0 for addition signs
if B0, then SC1 for (2ac2)3 = 8a3c6 or
72a5c7 seen
4
(i) (10, 4)
2
0 for (5, 4); otherwise 1 for each
coordinate
ignore accompanying working / description of
transformation;
condone omission of brackets;
(Image includes back page for examiners to check that
there is no work there)
4
(ii) (5, 11)
2
0 for (5, 4); otherwise 1 for each
coordinate
ignore accompanying working / description of
transformation;
condone omission of brackets
5
6000
4
M3 for 15 × 52 × 24;
condone inclusion of x4 eg (2x)4;
condone omission of brackets in 2x4 if 16 used;
or M2 for two of these elements correct
with multiplication or all three elements
correct but without multiplication (e.g.
in list or with addition signs);
allow M3 for correct term seen (often all terms written
down) but then wrong term evaluated or all evaluated
and correct term not identified;
or M1 for 15 soi or for 1 6 15 ... seen
in Pascal’s triangle;
15 × 52 × (2x)4 earns M3 even if followed by 15 × 25 ×
2 calculated;
SC2 for 20000[x3]
no MR for wrong power evaluated but SC for fourth
term evaluated
2
4751
6
Mark Scheme
3
2
2x + 9x + 4x – 15
3
as final answer; ignore ‘= 0’;
B2 for 3 correct terms of answer seen or
for an 8-term or 6 term expansion with
at most one error:
7
June 2011
correct 8-term expansion:
2x3 + 6x2 − 2x2 + 5x2 − 6x + 15x − 5x − 15
correct 6-term expansions:
2x3 + 4x2 + 5x2 − 6x + 10x − 15
2x3 + 6x2 + 3x2 + 9x − 5x − 15
2x3 + 11x2 − 2x2 + 15x − 11x − 15
or M1 for correct quadratic expansion
of one pair of brackets;
for M1, need not be simplified;
or SC1 for a quadratic expansion with
one error then a good attempt to
multiply by the remaining bracket
ie SC1 for knowing what to do and making a
reasonable attempt, even if an error at an early stage
means more marks not available
allow seen in formula; need not have numbers
substituted but discriminant part must be correct;
b 2 − 4ac soi
M1
1 www
A1
or B2
clearly found as discriminant, or stated as b 2 − 4ac , not
just seen in formula eg M1A0 for b 2 − 4ac = 1 = 1 ;
2 [distinct real roots]
B1
B0 for finding the roots but not saying
how many there are
condone discriminant not used; ignore incorrect roots
found
3
4751
8
Mark Scheme
yx + 3y = 1 − 2x oe or ft
M1
for multiplying to eliminate
denominator and for expanding
brackets,
or for correct division by y and writing
as separate fractions: x + 3 =
yx + 2x = 1 − 3y oe or ft
x (y + 2) = 1 − 3y oe or ft
[ x =]
1 − 3y
oe or ft as final answer
y+2
June 2011
1 2x
−
;
y y
each mark is for carrying out the operation correctly; ft
earlier errors for equivalent steps if error does not
simplify problem;
some common errors:
for collecting terms; dep on having an
ax term and an xy term, oe after division
by y,
y (x + 3) = 1 − 2x
yx + 3x = 1 − 2x M0
yx + 5x = 1 M1 ft
x (y + 5) = 1 M1 ft
M1
for taking out x factor; dep on having an
ax term and an xy term, oe after division
by y,
1
x=
y+5
M1
for division with no wrong work after;
dep on dividing by a two-term
expression; last M not earned for tripledecker fraction as final answer
M1
4
M1 ft
yx + 3 = 1 − 2x M0
yx + 2x = −2 M1 ft
x (y + 2) = −2 M1 ft
x=
−2
y+2
for M4, must be completely correct;
M1 ft
4751
9
Mark Scheme
June 2011
x + 2y = k (k ≠ 6) or
y = − ½ x + c (c ≠ 3)
M1
for attempt to use gradients of parallel
lines the same; M0 if just given line
used;
eg following an error in manipulation, getting original
line as y = ½ x + 3 then using y = ½ x + c earns M1 and
can then go on to get A0 for y = ½ x − 4, M1 for (0,
−4) M1 for (8, 0) and A0 for area of 16;
x + 2y = 12 or [y = ]− ½ x + 6 oe
A1
or B2; must be simplified; or evidence
of correct ‘stepping’ using (10, 1) eg
may be on diagram;
allow bod B2 for a candidate who goes straight to
y = − ½ x + 6 from 2y = −x + 6;
NB the equation of the line is not required; correct
intercepts obtained will imply this A1;
(12, 0) or ft
M1
or ‘when y = 0, x = 12’ etc
or using 12 or ft as a limit of
integration;
intersections must ft from their line or
‘stepping’ diagram using their gradient
(0, 6)or ft
M1
or integrating to give − ¼ x2 + 6x or ft
their line
36 [sq units] cao
A1
or B3 www
NB for intersections with axes, if both Ms are not
gained, it must be clear which coord is being found eg
M0 for intn with x axis = 6 from correct eqn;;
if the intersections are not explicit, they may be
implied by the area calculation eg use of ht = 6 or the
correct ft area found;
allow ft from the given line as well as others for both
these intersection Ms;
NB A0 if 36 is incorrectly obtained eg after
intersection x = −12 seen (which earns M0 from
correct line);
5
4751
10
Mark Scheme
n (n + 1)(n + 2)
M1
condone division by n and then
(n + 1)(n + 2) seen, or separate factors
shown after factor theorem used;
argument from general consecutive
numbers leading to:
June 2011
ignore ‘ = 0’;
an induction approach using the factors may also be
used eg by those doing paper FP1 as well;
at least one must be even
A1
[exactly] one must be multiple of 3
A1
or divisible by 2;
A0 for just substituting numbers for n and stating
results;
if M0:
allow SC1 for showing given
expression always even
6
allow SC2 for a correct induction approach using the
original cubic (SC1 for each of showing even and
showing divisible by 3)
4751
Mark Scheme
SECTION B
11 (i) x + 4x2 + 24x + 31 = 10 oe
4x2 + 25x + 21 [= 0]
(4x + 21)(x + 1)
11
M1
for subst of x or y or subtraction to
eliminate variable; condone one error;
M1
for collection of terms and
rearrangement to zero; condone one
error;
M1
x = −1 or −21/4 oe isw
A1
y = 11 or 61/4 oe isw
A1
(ii) 4(x + 3)2 – 5 isw
4
for factors giving at least two terms of
their quadratic correct or for subst into
formula with no more than two errors
[dependent on attempt to rearrange to
zero];
or A1 for (−1, 11) and A1 for (−21/4,
61/4) oe
June 2011
or 4y2 − 105y + 671 [= 0];
eg condone spurious y = 4x2 + 25x + 21 as one error
(and then count as eligible for 3rd M1);
or (y − 11)(4y − 61);
[for full use of completing square with no more than
two errors allow 2nd and 3rd M1s simultaneously];
from formula: accept x = −1 or −42/8 oe isw
B1 for a = 4,
eg an answer of (x + 3)2 – 5/ 4 earns B0 B1 M1;
B1 for b = 3,
B2 for c =−5 or M1 for 31 − 4 × their b2 1(2x + 6)2 − 5 earns B0 B0 B2;
soi or for −5/4 or for 31/4 − their b2 soi
4( earns first B1;
condone omission of square symbol
11
(iii)(A) x = −3 or ft (−their b) from (ii)
1
11
(iii)(B) −5 or ft their c from (ii)
1
0 for just −3 or ft;
0 for x = −3, y = −5 or ft
0 for just (−3, −5);
bod 1 for x = −3 stated then y = −5 or ft
allow y = −5 or ft
7
4751
12
12
Mark Scheme
(i) y = 2x + 5 drawn
M1
−2, −1.4 to −1.2, 0.7 to 0.85
A2
(ii) 4 = 2x3 + 5x2 or 2 x + 5 −
4
=0
x2
June 2011
condone unruled and some doubling;
tolerance: must pass within/touch at least two circles
on overlay; the line must be drawn long enough to
intersect curve at least twice;
A1 for two of these correct
condone coordinates or factors
B1
condone omission of final ‘= 0’;
and completion to given answer
f(−2) = −16 + 20 − 4 = 0
B1
or correct division / inspection showing
that x + 2 is factor;
use of x + 2 as factor in long division
of given cubic as far as 2x3 + 4x2 in
working
M1
or inspection or equating coefficients,
with at least two terms correct;
2x2 + x − 2 obtained
A1
[ x =]
−1 ± 12 − 4 × 2 × −2
oe
2× 2
−1 ± 17
oe isw
4
M1
may be set out in grid format
condone omission of + sign (eg in grid format)
dep on previous M1 earned; for attempt
at formula or full attempt at completing
square, using their other factor
A1
8
not more than two errors in formula / substitution /
completing square; allow even if their ‘factor’ has a
remainder shown in working;
M0 for just an attempt to factorise
4751
12
13
Mark Scheme
(iii)
4
= x + 2 or y = x + 2 soi
x2
M1
y = x + 2 drawn
A1
1 real root
A1
(i) [radius = ] 4
B1
[centre] (4, 2)
B1
eg is earned by correct line drawn
June 2011
condone intent for line; allow slightly out of tolerance;
condone unruled; need drawn for −1.5 ≤ x ≤ 1.2; to
pass through/touch relevant circle(s) on overlay
B0 for ± 4
condone omission of brackets
9
4751
13
Mark Scheme
(ii) (x − 4)2 + (− 2)2 = 16 oe
M1
for subst y = 0 in circle eqn;
June 2011
NB candidates may expand and rearrange eqn first,
making errors – they can still earn this M1 when they
subst y = 0 in their circle eqn;
condone omission of (−2)2 for this first M1 only; not
for second and third M1s;
do not allow substitution of x = 0 for any Ms in this
part
(x − 4)2 = 12 or x2 − 8x + 4 [= 0]
M1
putting in form ready to solve by comp
sq, or for rearrangement to zero;
condone one error;
eg allow M1 for x2 + 4 = 0 [but this two-term quadratic
is not eligible for 3rd M1];
x − 4 = ± 12 or
M1
for attempt at comp square or formula;
dep on previous M2 earned and on
three-term quadratic;
not more than two errors in formula / substitution;
allow M1 for x − 4 = 12 ;
M0 for just an attempt to factorise
[ x =]
8 ± 82 − 4 × 1 × 4
2 ×1
[ x =]4 ± 12 or 4 ± 2 3 or
8 ± 48
oe
2
A1
isw
or
or
sketch showing centre (4, 2) and
triangle with hyp 4 and ht 2
M1
42 − 22 = 12
M1
or the square root of this;
implies previous M1 if no sketch seen;
[ x =]4 ± 12 oe
A2
A1 for one solution
10
4751
13
(
)
(iii) subst 4 + 2 2, 2 + 2 2 into
Mark Scheme
B1
circle eqn and showing at least one
step in correct completion
or showing sketch of centre C and A
and using Pythag:
(2 2 ) + (2 2 )
2
Sketch of both tangents
M1
grad tgt = −1 or −1/their grad CA
M1
2
June 2011
or subst the value for one coord in circle eqn and
correctly working out the other as a possible value;
= 8 + 8 = 16 ;
need not be ruled;
must have negative gradients with tangents intended to
be parallel and one touching above and to right of
centre; mark intent to touch – allow just missing or just
crossing circle twice; condone A not labelled
allow ft after correct method seen for
grad CA =
2+2 2 −2
4+2 2 −4
allow ft from wrong centre found in (i);
oe (may be on/
near sketch);
y − (2 + 2 2) = their m( x − (4 + 2 2))
M1
or y = their mx + c and subst of
4 + 2 2, 2 + 2 2 ;
for intent; condone lack of brackets for M1;
independent of previous Ms; condone grad of CA used;
y = − x + 6 + 4 2 oe isw
A1
accept simplified equivs eg
x+ y =6+4 2 ;
A0 if obtained as eqn of other tangent instead of the
tangent at A (eg after omission of brackets);
parallel tgt goes through
4 − 2 2, 2 − 2 2
M1
or ft wrong centre;
may be shown on diagram;
may be implied by correct equation for
the tangent (allow ft their gradient);
no bod for just y − 2 − 2 2 = −1( x − 4 − 2 2) without
first seeing correct coordinates;
eqn is y = − x + 6 − 4 2 oe isw
A1
(
)
(
)
accept simplified equivs eg
x+ y =6−4 2
Section B Total: 36
11
A0 if this is given as eqn of the tangent at A instead of
other tangent (eg after omission of brackets)
OCR (Oxford Cambridge and RSA Examinations)
1 Hills Road
Cambridge
CB1 2EU
OCR Customer Contact Centre
14 – 19 Qualifications (General)
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
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is a Company Limited by Guarantee
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Registered Office; 1 Hills Road, Cambridge, CB1 2EU
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Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2011
4751
Mark Scheme
Question
Answer
y = 2x + 7 isw
1
(0, 7) and (3.5, 0) oe or ft their y = 2x + c
Marks
2
June 2012
Guidance
M1 for y  1 = 2(x  3) or
1 = 2 × 3 + c oe
1
condone lack of brackets and eg y = 7,
x = 3.5 or ft isw but 0 for poor notation
such as (3.5, 7) and no better answers
seen
[3]
2
[b  ] 
3a
oe www
2c
3
M2 for [b 2 ]
3a
soi
2c
or M1 for other [b 2 ]
ka
a
or [b 2 ]
oe
c
kc
allow M1 for a triple-decker or
quadruple-decker fraction or decimals
eg
and M1 for correctly taking the square root
of their b2, including the ± sign;
3
3
(i)
(ii)
25
4
9
[3]
2
[2]
2
M1 for
1
1
25
4
x3
5
as final answer www
or 1 
x2
x2
1
2
2
or or   or
9
3
3
3
64
729
B2 for correct answer seen and then spoilt
M1 for (x + 3)(x  3)
and M1 for (x + 2)(x + 3)
5
square root must extend below the
fraction line
1
2
M1 for 4 or 9 or
[3]
1.5a
, if no recovery later
c
25
 1 
or   or 52 or
1
 25 
seen
[2]
3
3a
2c
eg M2 for [b ]
1
 64  3
0 for just 

 729 
4751
5
Mark Scheme
Question
(i)
Answer
30
Marks
3
June 2012
Guidance
M1 for
M1 for
 6
3
24  2 6 soi
or allow SC2 for final answer of 5
5
(ii)
8
11
[3]
2
M0 for 6000 6 ie cubing 10 as well
 6 6 soi and
 6
for those using indices:
M1 for both 10 × 63/2 and 2 × 61/2 oe
then M1 for 5 × 6 oe
2
or 5 36 or 10 9 etc
award SC2 for similar correct answer
with no denominator
M1 for common denominator
4  5 4  5 soi - may be in separate
condone lack of brackets



fractions
or for a final answer with denominator 11,
even if worked with only one fraction
6
(i)
10 cao
6
(ii)
720 [x3]
[2]
1
[1]
4
B3 for 720 [x3] or for 10 × 9 × 8 [x3]
or M2 for 10 × 32 × (2)3 oe or ft from (i)
or M1 for two of these three elements correct
or ft;
condone x still included
condone 720 x etc
allow equivalent marks for the x3 term
as part of a longer expansion
3

 2  
eg M2 for 35  ...10    ...  or M1

 3  

3
 2 
for 10    etc
 3 
[4]
6
4751
Question
7
Mark Scheme
Answer
4k  4 × 1 × 5 or k2  5 [< 0] oe
or [(x + k)2 +] 5  k2 [> 0] oe
2
Marks
M2
June 2012
Guidance
allow M2 for 2k2 < 20, 2k2  20 = 0 etc
but M1 only for just 2k2  20
2
or M1 for b − 4ac soi (may be in formula)
or for attempt at completing square
ignore rest of quadratic formula
allow =, > , ≤ etc instead of <
ignore b 2  4ac  0 seen if
b 2  4ac  0 then used, otherwise just
M1 for
8
 5k 5
A2
16 + 2b + c = 0 oe
[4]
M1
81  3b + c = 85 oe
B2
b 2  4 ac  0
may be two separate inequalities
or A1 for one ‘end’ correct
or B1 for ‘endpoint’ = 5
allow SC1 for  10  k  10
following at least M1 for 2k2  20 oe
need not be simplified; condone 8 or 32 as
first term if 24 not seen
in this question use annotation to
indicate where part marks are earned
M1 for f(3) seen or used, condoning one
error except +3b – need not be simplified
or for long division as far as obtaining
x3  3x2 in quotient
eg M1 for 81  3b + c = 0
‘long division’ may be seen in grid or
a mixture of methods may be used
eg B2 for c  3(b  27) = 85
20 + 5b = 0 oe
M1
for elimination of one variable, ft their
equations in b and c, condoning one error in
rearrangement of their original equations or
in one term in the elimination
correct operation must be used in
elimination
b = 4 and c = 8
A1
allow correct answers to imply last M1 after
correct earlier equations
for misread of x4 as x3 or x2 or higher
powers, allow all 3 Ms equivalently
[5]
7
4751
Question
9
Mark Scheme
Answer
6n + 9 isw or 3(2n + 3)
Marks
B1
6n is even [but 9 is odd], even + odd = odd
or
2n + 3 is odd since even + odd = odd and
odd × odd = odd
B1 dep
‘n is a multiple of 3’ or ‘n is divisible by 3’
without additional incorrect statement(s)
B2
June 2012
Guidance
this mark is dependent on the previous B1
accept equiv. general statements using either
6n + 9 or 3(2n + 3)
B2 for ‘it is divisible by 9, so n is divisible
by 3’
M1 for ‘6n is divisible by 9’ or ‘2n + 3 is
divisible by 3’ or for ‘n is a multiple of 3’ oe
with additional incorrect statement(s)
B2 for just ‘it is divisible by 3’
but M1 for ‘it is divisible by 9, so it is
divisible by 3’
eg M1 for ‘n is divisible by 9, so n is
divisible by 3’
N.B. 0 for ‘n is a factor of 3’ (but M1
may be earned earlier)
[4]
8
4751
Question
10 (i)
Mark Scheme
Answer
AB = (1(1)) + (5  1)2
2
2
Marks
M1
June 2012
Guidance
oe, or square root of this; condone poor
notation re roots; condone (1 + 1)2 instead of
(1(1))2
 2 
allow M1 for vector AB =   , condoning
 4 
poor notation, or triangle with hyp AB and
lengths 2 and 4 correctly marked
BC2 = (3  (1))2 + (1 1)2
M1
oe, or square root of this; condone poor
notation re roots; condone (3 + 1)2 instead of
(3(1))2 oe
 4
allow M1 for vector BC =   , condoning
 2 
poor notation, or triangle with hyp BC and
lengths 4 and 2 correctly marked
shown equal eg
AB2 = 22 + 42 [=20] and
BC2 = 42 + 22 [=20] with correct notation for
final comparison
A1
or statement that AB and BC are each the
hypotenuse of a right-angled triangle with
sides 2 and 4 so are equal
SC2 for just AB2 = 22 + 42 and
BC2 = 42 + 22 (or roots of these) with no
clearer earlier working; condone poor
notation
[3]
9
eg A0 for AB = 20 etc
4751
Question
10 (ii)
Mark Scheme
Answer
5   1
6
[grad. of AC =]
oe
or
2
1 3
5 1
4
[grad. of BD =]
oe
or
11   1
12
showing or stating product of gradients = 1
or that one gradient is the negative reciprocal
of the other oe
Marks
M1
Guidance
award at first step shown even if errors after
M1
B1
June 2012
if one or both of grad AC = -3 and
grad BD = 1/3 seen without better
working for both gradients, award one
M1 only. For M1M1 it must be clear
that they are obtained independently
eg accept m1 × m2 = 1 or ‘one gradient is
negative reciprocal of the other’
B0 for ‘opposite’ used instead of ‘negative’
or ‘reciprocal’
[3]
10
may be earned independently of
correct gradients, but for all 3 marks to
be earned the work must be fully
correct
4751
Question
10 (iii)
Mark Scheme
Answer
midpoint E of AC = (2, 2) www
eqn BD is y  13 x  34 oe
Marks
B1
M1
Guidance
condone missing brackets for both B1s
0 for ((5+ 1)/2, (1+3)/2) = (2, 2)
accept any correct form isw or correct ft their
gradients or their midpt F of BD
this mark will often be gained on the first
line of their working for BD
eqn AC is y = 3x + 8 oe
M1
June 2012
accept any correct form isw or correct ft their
gradients or their midpt E of AC
may be earned using (2, 2) but then
must independently show that B or D
or (5, 3) is on this line to be eligible for
A1
if equation(s) of lines are seen in part
ii, allow the M1s if seen/used in this
part
this mark will often be gained on the first
line of their working for AC
[see appendix for alternative methods instead
showing E is on BD for this M1]
using both lines and obtaining intersection E
is (2, 2) (NB must be independently obtained
from midpt of AC)
A1
midpoint F of BD = (5,3)
B1
[see appendix for alternative ways of
gaining these last two marks in
different methods]
this mark is often earned earlier
see the appendix for some common
alternative methods for this question; for all
methods, for A1 to be earned, all work for
the 5 marks must be correct
[5]
11
for all methods show annotations M1
B1 etc then omission mark or A0 if
that mark has not been earned
4751
Question
11 (i)
Mark Scheme
Answer
(2x + 1)(x + 2)(x  5)
Marks
M1
correct expansion of two linear factors of
their product of three linear factors
M1
expansion of their linear and quadratic
factors
M1
or (x + 1/2)(x + 2)(x  5);
need not be written as product
June 2012
Guidance
throughout, ignore ‘=0’
for all Ms in this part condone missing
brackets if used correctly
dep on first M1; ft one error in previous
expansion; condone one error in this
expansion
or for direct expansion of all three factors,
allow M2 for
2x3  10x2+ 4x2 + x2  20x  5x + 2x  10 [or
half all these], or M1 if one or two errors,
[y =] 2x3  5x2  23x  10 or a = 5, b =  23
and c =  10
A1
dep on first M1
condone poor notation when
‘doubling’ to reach expression with
2x3...
for an attempt at setting up three
simultaneous equations in a, b, and c:
M1 for at least two of the three equations
then M2 for correctly eliminating any two
variables or M1 for correctly eliminating one
variable to get two equations in two
unknowns
and then A1 for values.
[4]
12
250 + 25a + 5b + c = 0
16 + 4a 2b + c = 0
¼+ ¼ a  ½ b + c = 0 oe
4751
Mark Scheme
Question
11 (ii)
Answer
graph of cubic correct way up
crossing x axis at 2, 1/2 and 5
Marks
B1
B1
June 2012
Guidance
must not be ruled; no curving back;
condone slight ‘flicking out’ at ends;
allow min on y axis or in 3rd or 4th
quadrants; condone some ‘doubling’ or
‘feathering’ (deleted work still may
show in scans)
B0 if stops at x-axis
on graph or nearby in this part
allow if no graph, but marked on xaxis
mark intent for intersections with both axes
crossing y axis at 10 or ft their cubic in (i)
B1
or x = 0, y = −10 or ft in this part if
consistent with graph drawn;
allow if no graph, but eg B0 for graph
nowhere near their indicated 10 or ft
11
(iii)
(0, 18); accept 18 or ft their constant  8
11
(iv)
roots at 2.5, 1, 8
[3]
1
[1]
M1
(2x  5)(x  1)(x  8)
A1
accept 2(x  2.5) oe instead of (2x  5)
M0 for use of (x + 3) or roots 3.5, 5,
2 but then allow
SC1 for (2x + 7)(x + 5)(x  2)
(0, 40); accept 40
B2
M1 for 5 × 1 × 8 or ft or for f(3)
attempted or g(0) attempted or for their
answer ft from their factorised form
eg M1 for (0, 70) or 70 after
(2x + 7)(x + 5)(x  2)
after M0, allow SC1 for f(3) = 70
or ft their intn on y-axis  8
or attempt to substitute (x  3) in
(2x + 1)(x + 2)(x  5) or in
(x + 1/2)(x + 2)(x  5) or in their unfactorised
form of f(x)– attempt need not be simplified
[4]
13
4751
Mark Scheme
Question
12 (i)
Answer
(1, 6) (0,1) (1,2) (2,3) (3,2) (4, 1) (5,6)
seen plotted
Marks
B2
June 2012
Guidance
or for a curve within 2 mm of these points;
use overlay; scroll down to spare copy
B1 for 3 correct plots or for at least 3 of the
of graph to see if used [or click ‘fit
pairs of values seen eg in table
height’
also allow B1 for ( 2  3 , 0) and
(2, 3) seen or plotted and curve not
through other correct points
12
(ii)
smooth curve through all 7 points
B1 dep
(0.3 to 0.5, 0.3 to 0.5) and
(2.5 to 2.7, 2.5 to 2.7) and (4, 1)
B2
1
 x2  4 x  1
x3
1 = (x  3)( x2  4x + 1)
dep on correct points; tolerance 2 mm;
condone some feathering/ doubling
(deleted work still may show in scans);
curve should not be flat-bottomed or
go to a point at min. or curve back in at
top;
may be given in form x = ..., y = ...
B1 for two intersections correct or for all the
x values given correctly
[5]
M1
M1
condone omission of brackets only if used
correctly afterwards, with at most one error;
condone omission of ‘=1’ for this M1
only if it reappears
allow for terms expanded correctly
with at most one error
at least one further correct interim step with
‘=1’ or ‘=0’ ,as appropriate, leading to given
answer, which must be stated correctly
A1
there may also be a previous step of
expansion of terms without an equation, eg
in grid
if M0, allow SC1 for correct division of
given cubic by quadratic to gain (x  3) with
remainder 1, or vice-versa
[3]
14
NB mark method not answer given answer is x3  7x2 + 13x  4 = 0
4751
Question
12 (iii)
Mark Scheme
Answer
quadratic factor is
x2  3x + 1
Marks
B2
June 2012
Guidance
found by division or inspection;
allow M1 for division by x  4 as far as
x3  4x2 in the working, or for inspection
with two terms correct
substitution into quadratic formula or for
completing the square used as far as
2
 x  32   54
M1
condone one error
no ft from a wrong ‘factor’;
3 5
oe
2
A2
A1 if one error in final numerical expression,
but only if roots are real
isw factors
[5]
Appendix: alternative methods for 10(iii) [details of equations etc are in main scheme]
for a mixture of methods, look for the method which gives most benefit to candidate, but take care not to award the second M1 twice
the final A1 is not earned if there is wrong work leading to the required statements
ignore wrong working which has not been used for the required statements
for full marks to be earned in this part, there must be enough to show both the required statements
find midpt E of AC
find eqn BD
B1
M1
find midpt E of AC
find eqn BD
B1
M1
find midpt E of AC
find eqn BD
B1
M1
find midpt E of AC
use gradients or vectors to
show E is on BD eg
grad BE = 2211  13 and grad
B1
M2
ED = 11522  13
[condone poor vector
notation]
show E on BD
find midpt F of BD
M1
B1
show E on BD
find midpt F of BD
M1
B1
state so not E
A1
find eqn of AC and correctly
show F not on AC (the
correct eqn for AC earns the
second M1 as per the main
scheme, if not already
earned)
A1
show E on BD
show BE2 = 10 and DE2 =
90 oe
showing BE2 = 10 and DE2
= 90 oe earns this A mark
as well as the B1 if there are
no errors elsewhere
[5]
M1
B1
A1
find midpt F of BD
B1
state so not E or
show F not on AC
A1
5]
15
4751
Mark Scheme
Question
1
(i)
Answer
9
or 0.36 isw
Marks
2
25
January 2013
Guidance
M1 for numerator or denominator correct or
M1 for eg
for squaring correctly or for inverting
correctly
for
3
 
5
1
 25 


 9 
2
or
M0 for just
or
 25 


 9 
1
or
25
9
3
5
1
5
 
3
2
[2]
1
(ii)
27
2
1
M1 for
814  3
soi
eg M1 for 33
M0 for 813= 531441 (true but not
helpful)
[2]
2
4x4y-3 or
4x
y
3
4
as final answer
3
B1 each ‘term’;
or M1 for numerator = 64x15y3 and M1 for
denominator = 16x11y6
[3]
5
B0 if obtained fortuitously
mark B scheme or M scheme to
advantage of candidate, but not a
mixture of both schemes
or
4751
Question
3
Mark Scheme
Answer
obtaining a correct relationship in any 3 of C,
d, r and A
Marks
M2
January 2013
may substitute into given relationship;
Guidance
eg M2 for Cd = 4πr2
or πd2 = kπr2 seen/obtained
or M1 for at least two of A = πr2, C = πd,
or obtaining a correct relationship in k and no
more than 2 other variables
C = 2πr, d = 2r or
r 
d
seen or used
2
condone eg Area = πr2;
allow
r 
d
 d 
A   
 2 
2
to imply A = πr2 and
and so earn M1, if M2 not
2
earned
convincing argument leading to k = 4
A1
must be from general argument, not just
substituting values for r or d;
may start from given relationship and derive
k=4
eg M1only for eg A = πr2 and C = πd
and so k = 4 with no further evidence
for factors giving at least two out of three
terms correct when expanded and collected
or use of formula or completing the
square with at most one error (comp
square must reach [5](x  a)2 ≤ b oe or
(5x  c)2 ≤ d oe stage)
if correct: 5(x  2.8)2 ≤ 51.2 or
(x  2.8)2 ≤ 10.24 or (5x  14)2 ≤ 256
[3]
4
(5x + 2)(x  6)
M1
boundary values 0.4 oe and 6 soi
A1
A0 for just
28 
1024
10
0.4 ≤ x ≤ 6 oe
A2
may be separate inequalities; mark final
answer
condone unsimplified but correct
28 
1024
10
 x 
28 
1024
etc
10
A1 for one end correct eg x ≤ 6
or for 0.4 < x < 6 oe
allow A1 for 0.4 ≤ 0 ≤ 6
or B1 for a ≤ x ≤ b ft their boundary values
condone errors in the inequality signs
during working towards final answer
[4]
6
4751
Question
5
Mark Scheme
Answer
4 + 2k + c = 0 or 22 + 2k + c = 0
Marks
B1
January 2013
Guidance
may be rearranged
9  3k + c = 35
B1
may be rearranged; the (3)2 must be
evaluated / used as 9
condone 32 seen if used as 9
correct method to eliminate one variable from
their eqns
M1
eg subtraction or substitution for c; condone
one error
M0 for addition of eqns unless also
multiplied appropriately
k = 6, c = 8
A1
from fully correct method, allowing recovery
from slips
if no errors and no method seen, allow
correct answers to imply M1 provided
B1B1 has been earned
or
[x2 + kx + c =] (x  2)(x  a)
or
M1
 5 × (3  a) = 35 oe
M1
a=4
k = 6, c = 8
A1
A1
or (x  2)(x + b)
[4]
7
4751
Mark Scheme
Question
Answer
6
identifying term as
3  5 
20  2 x   
 x 
Marks
M3
3
January 2013
Guidance
xs may be omitted; eg M3 for
20 × 8 × 125
condone lack of brackets;
oe
M1 for  k   2 x 
3
 5 
 
 x 
3
soi (eg in list or table),
condoning lack of brackets
and M1 for k = 20 or eg
65 4
first M1 not earned if elements added
not multiplied; otherwise, if in list or
table bod intent to multiply
M0 for binomial coefficient if it still
has factorial notation
3 21
or for 1 6 15 20 15 6 1 seen (eg Pascal’s
triangle seen, even if no attempt at expansion)
and M1 for selecting the appropriate term (eg
may be implied by use of only k = 20, but this
M1 is not dependent on the correct k used)
20 000
A1
may be gained even if elements added
or B4 for 20 000 obtained from multiplying
out
5 

2x  
x 

6
allow SC3 for 20000 as part of an expansion
[4]
7
(i)
9
3
www oe as final answer
2
M1 for
48  4
3
or
75  5
3
soi
[2]
7
(ii)
39  7
5
3
www as final answer
44
M1 for attempt to multiply numerator and
denominator by 7  5
B1 for each of numerator and denominator
correct (must be simplified)
[3]
8
condone
39
44

7
5
for 3 marks
44
eg M0B1 if denominator correctly
rationalised to 44 but numerator not
multiplied
4751
Mark Scheme
Question
Answer
5c + 9t = 2ac + at
8
Marks
M1
January 2013
Guidance
for correct expansion of brackets
5c  2ac = at  9t oe
M1
for correct collection of terms, ft
eg after M0 for 5c + 9t = 2ac + t allow this
M1 for 5c  2ac = 8t oe
for each M, ft previous errors if their
eqn is of similar difficulty;
c(5  2a) = at  9t oe
M1
for correctly factorising, ft; must be
c × a two-term factor
may be earned before t terms collected
M1
for correct division, ft their two-term factor
treat as MR if t is the subject, with a
penalty of 1 mark from those gained,
marking similarly
c  
at  9t
5  2a
or
t(a  9)
5  2a
oe as final answer
[4]
9
(i)
sketch of cubic the right way up, with two tps
B1
their graph touching the x-axis at 2 and
crossing it at 3 and no other places
B1
intersection of y-axis at 12
B1
No section to be ruled; no curving
back; condone some curving out at
ends but not approaching another
turning point; condone some doubling
(eg erased curves may continue to
show); ignore position of turning points
for this mark
if intns are not labelled, they must be shown
nearby
mark intent if ‘daylight’ between curve
and axis at x = 2
if no graph but 12 marked on y-axis,
or in table, allow this 3rd mark
[3]
9
(ii)
5 and 0
B2
B1 each; allow B2 for 5, 5, 0;
or B1 for both correct with one extra value
or for (5, 0) and (0, 0)
or SC1 for both of 1 and 6
[2]
9
if their graph wrong, allow 5 and 0
from starting again with eqn, or ft their
graph with two intns with x-axis
4751
10
Question
(i)
Mark Scheme
Answer
midpt of AB =
 1 5 
 , 
 2 2 
oe www
Marks
B2
January 2013
allow unsimplified
B1 for one coordinate correct
Guidance
if working shown, should come from
 3  2 4  1 
,


2
2 

oe
NB B0 for x coord. =
5
, (obtained
2
from subtraction instead of addition)
4 1
grad AB =
3  (2)
oe
M1
must be obtained independently of given line;
accept 3 and 5 correctly shown eg in a sketch,
followed by 3/5
M1 for rise/run = 3/5 etc
M0 for just 3/5 with no evidence
using gradient of AB to obtain grad perp
bisector
M1
for use of m1m2 = 1 soi or ft their gradient
AB
for those who find eqn of AB first, M0
for just
y  4 
y  4
1 4
1 4
2  3

x 3
2  3
x
 3
oe, but M1 for
oe
ignore their going on to find the eqn of
AB after finding grad AB
this second M1 available for starting
with given line =
5
and obtaining
3
M0 for just
5
without AB grad found
grad. of AB from it
3
y  2.5 =
5
(x  0.5) oe
M1
eg M1 for
y 
5
x  c
and subst of midpt;
3
3
ft their gradient of perp bisector and midpt;
M0 for just rearranging given equation
10
no ft for gradient of AB used
4751
Mark Scheme
Question
Answer
completion to given answer 3y + 5x = 10,
showing at least one interim step
Marks
M1
January 2013
Guidance
condone a slight slip if they recover quickly
NB answer given; mark process not
and general steps are correct ( eg sometimes a answer; annotate if full marks not
earned eg with a tick for each mark
5
x  c
slip in working with the c in y 
earned
3
- condone 3y = 5x + c followed by
substitution and consistent working)
scores such as B2M0M0M1M1 are
possible
M0 if clearly ‘fudging'
after B2, allow full marks for complete
method of showing given line has
gradient perp to AB (grad AB must be
found independently at some stage) and
passes through midpt of AB
[6]
10
(ii)
3y + 5(4y  21) = 10
M1
or other valid strategy for eliminating one
variable attempted eg
5
3
x 
10
3

x

4
21
;
4
condone one error
or eg 20y = 5x + 105 and subtraction of
two eqns attempted
no ft from wrong perp bisector eqn,
since given
allow M1 for candidates who reach
y = 115/23 and then make a worse
attempt, thinking they have gone wrong
(1, 5) or y = 5, x = 1 isw
A2
A1 for each value;
if AO allow SC1 for both values correct but
unsimplified fractions, eg
[3]
11
 23 115 
,


23 
 23
NB M0A0 in this part for finding E
using info from (iii) that implies E is
midpt of CD
4751
10
Question
(iii)
Mark Scheme
Answer
(x  a) + (y  b) = r2 seen or used
2
2
12 + 42 oe (may be unsimplified), from clear
use of A or B
Marks
M1
M1
January 2013
Guidance
or for (x + 1) + (y  5) =k, or ft their E,
where k > 0
2
2
for calculating AE or BE or their squares, or
for subst coords of A or B into circle eqn to
find r or r2, ft their E;
this M not earned for use of CE or DE
or ½ CD
NB some cands finding AB2 = 34 then
obtaining 17 erroneously so M0
(x + 1)2 + (y  5)2 = 17
A1
for eqn of circle centre E, through A and B;
allow A1 for r2 = 17 found after
(x + 1)2 + (y  5)2 = r2 stated and second M1
clearly earned
if (x + 1)2 + (y  5)2 = 17 appears without
clear evidence of using A or B, allow the first
M1 then M0 SC1
showing midpt of CD = (1, 5)
M1
showing CE or DE =
of C and D on circle
M1
17
oe or showing one
alt M1 for showing CD2 = 68 oe
allow to be earned earlier as an invalid
attempt to find r
12
SC also earned if circle comes from C
or D and E, but may recover and earn
the second M1 later by using A or B
4751
Mark Scheme
Question
Answer
Marks
January 2013
Guidance
showing that both C and D are on circle and
other methods exist, eg: may find eqn
commenting that E is on CD is enough for
of circle with centre E and through C or
last M1M1;
D and then show that A and B and
similarly showing CD2 = 68 and both C and
other of C/D are on this circle – the
D are on circle oe earns last M1M1
marks are then earned in a different
order; award M1 for first fact shown
and then final M1 for completing the
argument;
if part-marks earned, annotate with a
tick for each mark earned beside where
earned
[5]
11
(i)
B3
2
5 
1

x   
2
4

oe
B1 for a = 5/2 oe
and M1 for 6  their a2 soi;
2
condone
5 
1

x   
2
4

oe = 0
condone omission of index –can earn
all marks
bod M1 for 6  4.25 or 6  25/2 etc, if
bearing some relation to an attempt at
6  their 2.52; M0 for just 1.75 etc
without further evidence
1 
 5
 , 
4
 2
oe or ft
B1
accept x = 2.5, y = 0.25 oe
[4]
13
condone starting again and finding
using calculus
4751
11
Mark Scheme
Question
(ii)
Answer
(2, 0) and (3, 0)
Marks
B2
B1 each
January 2013
Guidance
condone not expressed as coordinates,
for both x and y values;
or B1 for both correct plus an extra
or M1 for (x  2)(x  3) or correct use of
formula or for th e ir a  th e ir b ft from (i)
(0, 6)
B1
graph of quadratic the correct way up and
crossing both axes
B1
accept eg in table or marked on graph
ignore label of their tp;
condone stopping at y-axis
condone ‘U’ shape or slight curving
back in/out; condone some doubling /
feathering – deleted work sometimes
still shows up in scoris; must not be
ruled; condone fairly straight with clear
attempt at curve at minimum; be
reasonably generous on attempt at
symmetry
[4]
11
(iii)
x2  5x + 6 = 2  x
M1
for attempt to equate or subtract eqns
or attempt at rearrangement and elimination
of x
accept calculus approach:
y = 2x  5
x2  4x + 4 [= 0]
M1
for rearrangement to zero ft and collection of
terms; condone one error;
if using completing the square, need to get as
far as (x  k)2 = c, with at most one error
[(x  2)2 = 0 if correct]
use of y = 1 M1
14
4751
Mark Scheme
Question
Answer
x = 2, [y = 0]
Marks
A1
January 2013
Guidance
condone omission of y = 0 since already
x = 2 A1
found in (ii)
if they have eliminated x, y = 0 is not sufft for
A1 – need to get x = 2
A0 for x = 2 and another root
‘double root at x = 2 so tangent’ oe; www;
A1
eg ‘only one point of contact, so tangent’;
or showing b2  4ac = 0, and concluding ‘so
tangent’; www
tgt is y [ 0] = (x  2) and obtaining
given line A1
[4]
12
(i)
f(1) = 11 +1 + 9  10 [= 0]
B1
allow for correct division of f(x) by (x  1)
showing there is no remainder,
condone 14 – 13 + 12 + 9 – 10
or for (x − 1)( x3 + x + 10) found, showing it
‘works’ by multiplying it out
attempt at division by (x − 1) as far as x4− x3
in working
M1
allow equiv for (x + 2) as far as
working
x
4
 2x
3
in
eg for inspection, M1 for two terms
right and two wrong
or for inspection with at least two terms of
cubic factor correct
correctly obtaining x3 + x + 10
A1
or x3  3x2 + 7x  5
[3]
15
if M0 and this division / factorising is
done in part (ii) or (iii), allow SC1 if
correct cubic obtained there; attach the
relevant part to (i) with a formal chain
link if not already seen in the image
zone for (i)
4751
12
Question
(ii)
Mark Scheme
Answer
[g(2) =] 8  2 + 10
or f(2) = 16 + 8 + 4  18  10
x =  2 isw
Marks
M1
A1
January 2013
Guidance
[in this scheme g(x) = x3 + x + 10]
eg f(2) = 16  8 + 4 + 18  10 or 20
allow M1 for correct trials with at least two
f(3) = 81  27 + 9 + 27  10 or 80
values of x (other than 1) using g(x) or f(x) or f(0) =  10
x3  3x2 + 7x  5
f(1) = 1 + 1 + 1  9  10 or 16
(may allow similar correct trials using
division or inspection)
No ft from wrong cubic ‘factors’ from
(i)
allow these marks if already earned in (i)
NB factorising of x3 + x + 10 or
x3  3x2 + 7x  5 in (ii) earns credit for
(iii) [annotate with a yellow line in
both parts to alert you – the image zone
for (iii) includes part (ii)]
[2]
16
4751
Mark Scheme
Question
1
Answer
y = 0.5x + 3 oe www isw
Marks
3
June 2013
Guidance
for 3 marks must be in form y = ax + b
B2 for 2y = x + 6 oe
or M1 for gradient = 
1
oe seen or used
2
and M1 for y  1 = their m (x  4)
substitution to eliminate one variable
2
3
3
(i)
(ii)
[3]
M1
or multiplication to make one pair of
coefficients the same;
condone one error in either method
simplification to ax = b or ax  b = 0 form,
or equivalent for y
M1
or appropriate subtraction / addition;
condone one error in either method
(0.7, 0.1) oe or x = 0.7, y = 0.1 oe isw
A2
[4]
2
A1 each
25
8a9
[2]
3
or M1 for y = their mx + c and (4, 1)
substituted
2
2
 10 
 1 
M1 for   or 
 oe soi
 2
 0.2 
1
or for
oe
0.04
1
B2 for 8 or M1 for 16 4  2 soi
9
and B1 for a
[3]
5
independent of first M1
ie M1 for one of the two powers used
correctly
M0 for just
1
with no other working
0.4
ignore ±
eg M1 for 23; M0 for just 2
4751
4
5
Mark Scheme
r
3V
oe www as final answer
 ( a  b)
3
f(2) = 18 seen or used
[3]
M1
32 + 2k  20 = 18 oe
A1
[k =] 3
A1
[3]
June 2013
M1 for dealing correctly with 3
M0 if triple-decker fraction, at the stage
where it happens, then ft;
and M1 for dealing correctly with π(a + b), ft
condone missing bracket at rh end
and M1 for correctly finding square root, ft
their ‘r2 =’; square root symbol must extend
below the fraction line
M0 if ±... or r > ...
for M3, final answer must be correct
or long division oe as far as obtaining a
remainder (ie not involving x) and equating
that remainder to 18 (there may be errors
along the way)
after long division: 2(k + 16)  20 = 18 oe
6
A0 for just 25 instead of 32 unless 32
implied by further work
4751
Mark Scheme
2560 www
6
4
June 2013
B3 for 2560 from correct term (NB
coefficient of x4 is 2560)
ignore terms for other powers; condone
x3 included;
or B3 for neg answer following 10 × 4 × 64
and then an error in multiplication
but eg 10 × 4 × 64 = 40  64 = 24
gets M2 only
or M2 for 10 × 22 × (4)3 oe; must have multn
signs or be followed by a clear attempt at
multn;
condone missing brackets eg allow M2
for 10 × 22 × 4x3
5
C3 or factorial notation is not
5  4  3  2 1
oe
sufficient but accept
2 1 3  2 1
or M1 for 22 × (4)3 oe (condone missing
brackets) or for 10 used or for 1 5 10 10 5 1
seen
10 may be unsimplified, as above
for those who find the coefft of x2 instead:
allow M1 for 10 used or for 1 5 10 10 5 1
seen ; and a further SC1 if they get 1280,
similarly for finding coefficient of x4 as 2560
7
(i)
53.5 oe or k = 7/2 oe
[4]
2
3
M1 for 125 = 5 or
[2]
7
1
2
5  5 soi
M1 only for eg 10, 22 and 4x3 seen in
table with no multn signs or evidence
of attempt at multn
[lack of neg sign in the x2 or x4 terms
means that these are easier and so not
eligible for just a 1 mark MR penalty]
M0 for just answer of 53 with no
reference to 125
4751
7
Mark Scheme
(ii)
attempting to multiply numerator and
denominator of fraction by 1  2 5
June 2013
M1
some cands are incorporating the
10  7 5 into the fraction. The M1s
are available even if this is done
wrongly or if 10  7 5 is also
multiplied by 1  2 5
8
denominator = 19 soi
M1
83 5
A1
3(x  2)2  7 isw or a = 3, b = 2 c = 7 www
[3]
4
must be obtained correctly, but independent
of first M1
eg M1 for denominator of 19 with a
minus sign in front of whole expression
or with attempt to change signs in
numerator
B1 each for a = 3, b = 2 oe
condone omission of square symbol;
ignore ‘= 0’
and B2 for c = 7 oe
or M1 for   
7
or for 5  their a(their b)2
3
5
2
  their b  soi
3
B0 for (2, 7)
may be implied by their answer
or for
9
(i)
7 or ft
B1
3n isw
[5]
1
[1]
accept equivalent general explanation
8
may be obtained by starting again eg
with calculus
4751
9
Mark Scheme
(ii)
at least one of (n  1)2 and (n + 1)2 correctly
expanded
M1
3 n2 + 2
B1
comment eg 3n2 is always a multiple of 3 so
remainder after dividing by 3 is always 2
B1
June 2013
M0 for just n2 + 1 + n2 + n2 + 1
must be seen
accept even if no expansions / wrong
expansions seen
dep on previous B1
B0 for just saying that 2 is not divisible by 3
– must comment on 3n2 term as well
allow B1 for
10
(i)
[radius =]
20 or 2 5 isw
[centre =] (3, 2)
[3]
B1
SC: n, n + 1, n + 2 used similarly can
obtain first M1, and allow final B1 for
similar comment on 3n2 + 6n + 5
3n 2  2
2
 n2 
3
3
B0 for  20 oe
B1
condone lack of brackets with
coordinates, here and in other questions
[2]
9
4751
10
Mark Scheme
(ii)
substitution of x = 0 or y = 0 into circle
equation
M1
or use of Pythagoras with radius and a
coordinate of the centre eg 20  22 or h2 + 32
= 20 ft their centre and/or radius
June 2013
equation may be expanded first, and
may include an error
bod intent
allow M1 for (x  3)2 = 20 and/or
(y  2)2 = 20
(x  7)(x + 1) [=0]
M1
no ft from wrong quadratic; for factors giving
two terms correct, or formula or completing
square used with at most one error
completing square attempt must reach
at least (x  a)2 = b
following use of Pythagoras allow M1
for attempt to add 3 to [±]4
(7, 0) and (1, 0) isw
[ y ]
4
 4 
2
 4  1  (7)
2
A1
accept x = 7 or 1 (both required)
M1
no ft from wrong quadratic; for formula or
completing square used with at most one
error
oe
completing square attempt must reach
at least (y  a)2 = b
following use of Pythagoras allow M1
for attempt to add 2 to [±] 11
 0, 2 
 4  44 
11 or  0,
 isw
2



A1
accept y 
[5]
10
4  44
oe isw
2
annotation is required if part marks are
earned in this part: putting a tick for
each mark earned is sufficient
4751
10
Mark Scheme
(iii)
show both A and B are on circle
B1
explicit substitution in circle equation and at
least one stage of interim working required oe
(4, 5)
B2
B1 each
June 2013
or clear use of Pythagoras to show AC
and BC each = 20
 7 1 6  4 
or M1 for 
,

2 
 2
10
B2
from correct midpoint and centre used; B1 for
 10
may be a longer method finding length
of ½ AB and using Pythag. with radius;
M1 for (4  3)2 + (5  2)2 or 12 + 32 or ft their
centre and/or midpoint, or for the square root
of this
no ft if one coord of midpoint is same
as that of centre so that distance
formula/Pythag is not required eg
centre correct and midpt (3, 1)
annotation is required if part marks are
earned in this part: putting a tick for
each mark earned is sufficient
[5]
11
4751
11
11
Mark Scheme
(i)
(ii)
sketch of cubic the right way up, with two tps
and clearly crossing the x axis in 3 places
B1
crossing/reaching the x-axis at 4, 2 and 1.5
B1
intersection of y-axis at 24
B1
2, 0 and 7/2 oe isw or ft their intersections
[3]
2
June 2013
no section to be ruled; no curving
back; condone slight ‘flicking out’
at ends but not approaching another
turning point; condone some
doubling (eg erased curves may
continue to show); accept min tp on
y-axis or in 3rd or 4th quadrant; curve
must clearly extend beyond the x
axis at both ‘ends’
intersections must be shown correctly
labelled or worked out nearby; mark
intent
accept curve crossing axis halfway
between 1 and 2 if 3/2 not marked
NB to find 24 some are expanding
f(x) here, which gains M1 in iiiA. If
this is done, put a yellow line here and
by (iii)A to alert you; this image
appears again there
B1 for 2 correct or ft or for
(2, 0) (0, 0) and (3.5, 0)
or M1 for (x + 2) x (2x  7) oe
or SC1 for 6, 4 and 1/2 oe
[2]
12
4751
11
Mark Scheme
(iii)
(A)
correct expansion of product of 2 brackets of
f(x)
correct expansion of quadratic and linear and
completion to given answer
M1
A1
June 2013
need not be simplified; condone lack of
brackets for M1
eg 2x2 + 5x  12 or 2x2 + x  6
or x2 + 6x + 8
or allow M1 for expansion of all 3 brackets,
showing all terms, with at most one error:
2x3 + 4x2 + 8x2  3x2 + 16x  12x  6x – 24
may be seen in (i) – allow the M1; the
part (i) work appears at the foot of the
image for (iii)A, so mark this rather
than in (i)
for correct completion if all 3 brackets
already expanded, with some reference to
show why 24 changes to 9
condone lack of brackets if they have
gone on to expand correctly; condone
‘+15’ appearing at some stage
NB answer given; mark the whole
process
[2]
13
4751
11
Mark Scheme
(iii)
(B)
June 2013
g(1) = 2 + 9  2  9 [=0]
B1
allow this mark for (x  1) shown to be a
factor and a statement that this means that
x = 1 is a root [of g(x) = 0] oe
B0 for just g(1) = 2(1)3 + 9(1)2  2(1) 
9 [=0]
attempt at division by (x − 1) as far as
2x3 − 2x2 in working
M1
or inspection with at least two terms of
quadratic factor correct
M0 for division by x + 1 after g(1) = 0
unless further working such as g(1) =
0 shown, but this can go on to gain last
M1A1
correctly obtaining 2x2 + 11x + 9
A1
allow B2 for another linear factor found by
the factor theorem
NB mixture of methods may be seen in
this part – mark equivalently
eg three uses of factor theorem,
or two uses plus inspection to get last
factor;
factorising a correct quadratic factor
M1
for factors giving two terms correct;
allow M1 for (x + 1)(x + 18/4) oe after
1 and 18/4 oe correctly found by
formula
eg allow M1 for factorising 2x2 + 7x  9 after
division by x + 1
(2x + 9)(x + 1)(x  1) isw
A1
allow 2(x + 9/2)(x + 1)(x  1) oe;
dependent on 2nd M1 only;
condone omission of first factor found; ignore
‘= 0’ seen
[5]
14
SC alternative method for last 4 marks:
allow first M1A1 for (2x + 9)(x2  1)
and then second M1A1 for full
factorisation
4751
12
Mark Scheme
(i)
June 2013
y = 2x + 3 drawn accurately
M1
at least as far as intersecting curve twice
(1.6 to 1.7, 0.2 to 0.3)
B1
intersections may be in form x = ..., y = ...
(2.1 to 2.2, 7.2 to 7.4)
B1
ruled straight line and within 2mm of
(2, 7) and (1, 1)
if marking by parts and you see work
relevant to (ii), put a yellow line here
and in (ii) to alert you to look
12
(ii)
1
 2x  3
x2
[3]
M1
or attempt at elimination of x by
rearrangement and substitution
may be seen in (i) – allow marks; the
part (i) work appears at the foot of the
image for (ii) so show marks there
rather than in (i)
1 = (2x + 3)(x  2)
M1
condone lack of brackets
implies first M1 if that step not seen
1 = 2x2  x  6 oe
A1
for correct expansion; need not be simplified;
implies second M1 if that step not seen
1
after
 2 x  3 seen
x2
NB A0 for 2x2  x  7 = 0 without expansion
seen [given answer]
1  12  4  2  7
oe
2 2
M1
use of formula or completing square on given
equation, with at most one error
1  57
isw
4
A1
isw eg coordinates;
after completing square, accept
better
[5]
15
1
57

or
4
16
completing square attempt must reach
at least [2](x  a)2 = b or (2x  c)2 = d
stage oe with at most one error
4751
12
Mark Scheme
(iii)
June 2013
1
  x  k and attempt at rearrangement
x2
M1
x 2   k  2  x  2k  1  0
M1
for simplifying and rearranging to zero;
eg M1 bod for x 2   k  2  x  2k
condone one error;
2
collection of x terms with bracket not required or M1 for x  2kx  2k  1  0
b2  4ac = 0 oe seen or used
M1
= 0 may not be seen, but may be
implied by their final values of k
[k =] 0 or 4 as final answer, both required
A1
SC1 for 0 and 4 found if 3rd M1 not earned
(may or may not have earned first two Ms)
[4]
16
eg obtained graphically or using
calculus and/or final answer given as a
range
4751
Mark Scheme
June 2013
Appendix: revised tolerances for modified papers for visually impaired candidates (graph in 12(i) with 6mm squares)
12
(i)
y = 2x + 3 drawn accurately
M1
at least as far as intersecting curve twice
(1.6 to 1.8, 0.2 to 0.3)
B1
intersections may be in form x = ..., y = ...
(2.1 to 2.3, 7.1 to 7.4)
B1
ruled straight line and within 3 mm of
(2, 7) and (1, 1)
if marking by parts and you see work
relevant to (ii), put a yellow line here
and in (ii) to alert you to look
[3]
17